Let's lay out the blueprint for this answer. First, we're going to cover some handy properties of the gauge function. Second, we will dip into some algebraic topology to prove your conjecture true in the specific case where $A$ and $B$ are both the closed unit ball in $\Bbb{R}^n$. Third, and finally, we use this result to prove your conjecture true for more general convex bodies $A$ and $B$ in $\Bbb{R}^n$.
To match with standard notation in algebraic topology, we will use
$$D^n = \{x \in \Bbb{R}^n : \|x\| \le 1\}$$
to be the closed unit ball/disc in $\Bbb{R}^n$.
The gauge function
Definition. Suppose we have a convex set $C$ that contains $0$ in its interior. Define the gauge function:
$$\gamma_C(x) = \inf\{\lambda \ge 0 : x \in \lambda C\}.$$
Note that, since $0 \in \operatorname{int} C$, this function is well-defined everywhere, and $\gamma_C(x) \ge 0$ for all $x$.
Proposition 1 $\gamma_C$ is a sublinear, and hence convex, function.
Proof. Note that, for $\mu > 0$ and $x \in \Bbb{R}^n$,
\begin{align*}
\gamma_C(\mu x) &= \inf\{\lambda \ge 0 : \mu x \in \lambda C\} \\
&= \inf\{\lambda \ge 0 : x \in \mu^{-1} \lambda C\} \\
&= \mu\inf\{\mu^{-1} \lambda \ge 0 : x \in \mu^{-1} \lambda C\} \\
&= \mu\gamma_C(x).
\end{align*}
If $\mu = 0$, then $\gamma_C(\mu x) = \gamma_C(0) = 0$.
Further, if $x, y \in \Bbb{R}^n$, then
\begin{align*}
\gamma_C(x)+\gamma_C(y) &= \inf\{\lambda \ge 0 : x \in \lambda C\} + \inf\{\mu \ge 0 : y \in \mu C\} \\
&= \inf \{\lambda + \mu : \lambda \ge 0, \mu \ge 0, x \in \lambda C, y \in \mu C\} \\
&\ge \inf \{\lambda + \mu \ge 0 : x + y \in (\lambda + \mu)C\} \\
&= \gamma_C(x + y).
\end{align*}
Thus, $\gamma_C$ is a sublinear, hence convex function. As it has full domain, it is also continuous. $\square$
Proposition 2. Suppose $C$ is convex, bounded, and $0 \in \operatorname{int} C$. Then $\gamma_C(x) = 0 \iff x = 0$.
Proof. If $x = 0$, then $0 \in \{0\} = 0C$, hence $\gamma_C(x) = 0$.
Conversely, let us suppose $\gamma_C(x) = 0$. In particular, there exists some sequence $\lambda_m \ge 0$ converging to $0$ such that $x \in \lambda_m C$. Since $C$ is bounded, there exists some $M$ such that $C \subseteq MD^n$. Thus,
$$x \in \lambda_m C \subseteq \lambda_m M D^n$$
for all $m$. That is,
$$x \in \bigcap_{m \in \Bbb{N}} \lambda_m M D^n = \{0\},$$
hence $x = 0$. $\square$
Proposition 3. Suppose that $C$ is bounded, convex, and $0 \in \operatorname{int} C$. Then, for all $x \in \Bbb{R}^n$, we have $x \in \gamma_C(x)\operatorname{cl} C$ (which shows, in particular, the infimum in the definition of $\gamma_C$ is a minimum when $C$ is closed and bounded).
Proof. From Proposition 2, if $x = 0$, then $x \in \{0\} = 0 \operatorname{cl} C = \gamma_C(x) \operatorname{cl} C$. Otherwise, $x \neq 0$, and hence $\gamma_C(x) > 0$.
Suppose $\lambda_m > \gamma_C(x)$ converges to $\gamma_C(x)$ such that $x \in \lambda_m C$. Then, $\frac{x}{\lambda_m} \in C$. Thus $\frac{x}{\lambda_m} \to \frac{x}{\gamma_C(x)} \in \operatorname{cl} C$, hence $x \in \gamma_C(x) \operatorname{cl} C$, and the minimum is attained. $\square$
Proposition 4. Suppose that $C$ is bounded, convex, and $0 \in \operatorname{int} C$. Then
- $\{0\} = \gamma_C^{-1}\{0\}$
- $\operatorname{int} C = \gamma_C^{-1}[0, 1)$
- $\operatorname{cl} C = \gamma_C^{-1}[0, 1]$
- $\Bbb{R}^n \setminus \operatorname{cl} C = \gamma^{-1}(1, \infty)$
- $\partial C = \gamma_C^{-1}\{1\}$.
Proof. Part 1 is simply Proposition 2.
For part 2, suppose $\gamma_C(x) < 1$. If $\gamma_C(x) = 0$, then the previous part tells us that $x = 0 \in \operatorname{int} C$. Otherwise, by Proposition 3, $x \in \gamma_C(x) \operatorname{cl} C$, i.e. $\frac{x}{\gamma_C(x)} \in \operatorname {cl} C$. Since $\frac{1}{\gamma_C(x)} > 1$, we see $x$ lies on the interior of the line segment between $\frac{x}{\lambda}$ and $0 \in \operatorname{int} \operatorname {cl} C$. By a standard complex analysis result, this puts $x \in \operatorname{int} \operatorname {cl} C = \operatorname {int} C$.
On the other hand, if $x \in \operatorname{int} C$, then there exists some closed ball of radius $\varepsilon > 0$, centred at $x$, contained in $C$. We then know that
$$x + \varepsilon\frac{x}{\|x\|} = \left(1 + \frac{\varepsilon}{\|x\|}\right)x \in C,$$
hence
$$x \in \left(1 + \frac{\varepsilon}{\|x\|}\right)^{-1}C.$$
Clearly,
$$\left(1 + \frac{\varepsilon}{\|x\|}\right)^{-1} < 1,$$
so $\gamma_C(x) < 1$.
Part 3 initially follows in much the same vein as part 2, with a couple of minor exceptions. We still see that, when $\gamma_C(x) \le 1$, $x$ lies on the line segment between $0$ and $\frac{x}{\gamma_C(x)}$, although not necessarily in the interior. This places $x \in \operatorname{cl} C$.
Conversely, suppose $\gamma_C(x) \le 1$. As in part 2, if $\gamma_C(x) < 1$, then $x \in \operatorname{int} C \subseteq \operatorname{cl} C$. So, assume $\gamma_C(x) = 1$. It follows from Proposition 3 that $x \in \gamma_C(x) \operatorname{cl} C = \operatorname{cl} C$.
Parts 4 and 5 then follow immediately from the previous two parts. $\square$
Corollary 5. Suppose $C$ is as in Propositions 3 and 4, and $x \neq 0$. Then $\frac{x}{\gamma_C(x)} \in \partial C$.
Proof. By Proposition 1,
$$\gamma_C\left(\frac{x}{\gamma_C(x)}\right) = \frac{\gamma_C(x)}{\gamma_C(x)} = 1,$$
hence $x \in \partial C$ by Proposition 4.
Algebraic topology
Algebraic topology is not in my wheelhouse, which is why I was hoping someone else would respond to your question. I know enough to be able to apply some fundamental theorems, but not enough to prove them on request.
Definition. Suppose $A$ is a topological space, and $B \subseteq A$. Further, suppose $f : A \to A$ is continuous such that $f(A) = B$ and $f|_B$ is the identity map on $B$ (or, technically speaking, the inclusion map from $B$ into $A$). We say $f$ is a retraction map from $A$ onto $B$. We also say $B$ is a retract of $A$.
Proposition 6. Suppose $x_0 \in \operatorname{int} D^n$. Then $S^n$ is a retract of $D^n \setminus \{x_0\}$.
Proof. Let $C = D^n - x_0$, and define the continuous map $\alpha : D^n \setminus \{x_0\} \to D^n \setminus \{x_0\}$ by:
$$\alpha(x) = x_0 + \frac{x - x_0}{\gamma_C(x - x_0)}.$$
Note that $C$ is closed, bounded, convex and contains $0$ in the interior. Proposition 2 states that $\gamma_C(y) = 0$ if and only if $y = 0$, hence $\alpha$ is well-defined. Proposition 1 therefore shows $\alpha$ is continuous. By Corollary 5,
$$\frac{x - x_0}{\gamma_C(x - x_0)} \in \partial C = \partial D^n - x_0,$$
hence
$$\alpha(x) = x_0 + \frac{x - x_0}{\gamma_C(x - x_0)} \in \partial D^n.$$
Moreover, by Proposition 4, if $x \in \partial D^n$, then $x - x_0 \in \partial C$, so $\gamma_C(x - x_0) = 1$, hence $\alpha(x) = x$. Thus $\alpha$ is a retraction from $D^n \setminus \{x_0\}$ to $\partial D^n$. $\square$
The following is Corollary 2.15, found on page 114 of Algebraic Topology, by Allen Hatcher (which is freely available from his website). I offer no proof.
Theorem 7 [Hatcher, Corollary 2.15]. $\partial D^n$ is not a retract of $D^n$. Hence every map $f : D^n \to D^n$ has a fixed point.
We will now proceed to use this to prove a special case of your question, where $A = B = D^n$.
Proposition 8. Suppose $f : D^n \to D^n$ is continuous. Further, suppose $f$ stabilises and is bijective on $\partial D^n$. Then $f$ is surjective.
Proof. Suppose, for the sake of contradiction, that $f$ is not surjective. Then, there exists some $x_0 \in D^n \setminus f(D^n)$. Note that $x_0$ must lie in $\operatorname{int} D^n$.
Since $f|_{\partial D^n}$ is bijective and continuous with $\partial D^n$ compact, we know there exists a continuous inverse $g : \partial D^n \to \partial D^n$.
Using Proposition 6, let $\alpha$ be a retraction of $D^n \setminus \{x_0\}$ onto $\partial D^n$.
Define $\beta : D^n \to D^n$ by $\beta(x) = g(\alpha(f(x)))$. We will show that $\beta$ is a retraction from $D^n$ onto $\partial D^n$, contradicting Theorem 7, showing $f$ must indeed be surjective.
First, note that $\beta$ is well-defined: the range of $f$ is contained in $D^n \setminus \{x_0\}$, which is the domain of $\alpha$. The range of $\alpha$ is $\partial D^n$, which is the domain of $g$. Also, as all the component functions are continuous, so is their composition.
Second, if $x \in \partial D^n$, then $f(x) \in \partial D^n$, by assumption. Thus, $\alpha(f(x)) = f(x)$, so $\beta(x) = g(f(x)) = x$. Hence, $\beta$ fixes $\partial D^n$, proving $\beta$ is a retraction, as claimed, and the proposition is proven. $\square$
Your question, in general
We can use Proposition 8 to establish your conjecture more generally. We will, however, need one more map (I don't know if this map has a name, or even if it's a standard map).
Suppose that $A$ and $B$ are convex bodies in $\Bbb{R}^n$, each containing $0$ in their interiors. We will use the following function for the next couple of propositions:
$$\varphi_{A, B} : \Bbb{R}^n \to \Bbb{R}^n : x \mapsto \begin{cases} \frac{\gamma_A(x)}{\gamma_B(x)}x & \text{if } x \neq 0 \\ 0 & \text{if } x = 0. \end{cases}$$
Proposition 9. We have $\varphi_{A, B}$ is well-defined and continuous.
Proof. By Proposition 2, the map is well-defined. By proposition 1, it is continuous everywhere except possibly $0$, and so we must establish separately that it is continuous at $0$.
Since $B$ is bounded, we have $B \subseteq MD^n$ for some $M > 0$. Since $0 \in \operatorname{int} A$, there exists some $\xi > 0$ such that $\xi D^n \subseteq A$. Fix $\varepsilon > 0$, and let $\delta = \frac{\xi \varepsilon}{M} > 0$. Suppose that $\|x\| \le \delta$. We wish to show $\|\varphi_{A, B}(x)\| \le \varepsilon$, establishing continuity at $0$. Note that the $x = 0$ case is clear, so we assume $x \neq 0$.
Note that $\frac{Mx}{\varepsilon} \in \xi D^n \subseteq 1A$, thus $\gamma_A\left(\frac{Mx}{\varepsilon}\right) \le 1$, hence by Proposition 1, $\gamma_A(x) \le \frac{\varepsilon}{M}$.
On the other hand, by Corollary 5, $\frac{x}{\gamma_B(x)} \in B \subseteq M D^n$, hence $\|x\| \le M \gamma_B(x)$, i.e. $\gamma_B(x) \ge \frac{\|x\|}{M}$. Therefore
$$\frac{\gamma_A(x)}{\gamma_B(x)} \le \frac{\varepsilon}{M} \cdot \frac{M}{\|x\|} = \frac{\varepsilon}{\|x\|}.$$
Thus,
$$\|\varphi_{A, B}(x)\| = \left\|\frac{\gamma_A(x)}{\gamma_B(x)}x\right\| \le \|x\|\frac{\varepsilon}{\|x\|} = \varepsilon.$$
Thus, $\varphi_{A, B}$ is continuous everywhere. $\square$
Proposition 10. For all $x \in \Bbb{R}^n$ and $\lambda \ge 0$, $\varphi_{A, B}(\lambda x) = \lambda \varphi_{A, B}(x)$.
Proof. When $x = 0$ or $\lambda = 0$, the result is straightforward. Otherwise, by Proposition 1,
$$\varphi_{A, B}(\lambda x) = \frac{\gamma_A(\lambda x)}{\gamma_B(\lambda x)}(\lambda x) = \lambda \frac{\lambda \gamma_A(x)}{\lambda \gamma_B(x)} x = \lambda \varphi_{A, B}(x).$$
Corollary 11. $\varphi_{B, A} = \varphi_{A, B}^{-1}$, hence $\varphi_{A, B}$ is a homeomorphism.
Proof. It suffices to show $\varphi_{B, A} \circ \varphi_{A, B} = \operatorname{Id}_{\Bbb{R}^n}$ by swapping $A$ and $B$. Note that $\varphi_{B, A}(\varphi_{A, B}(0)) = \varphi_{B, A}(0) = 0$. Let us assume $x \in \Bbb{R}^n \setminus \{0\}$. Then, by Proposition 10,
$$\varphi_{B, A}(\varphi_{A, B}(x)) = \varphi_{B, A}\left(\frac{\gamma_A(x)}{\gamma_B(x)}x\right) = \frac{\gamma_A(x)}{\gamma_B(x)} \varphi_{B, A}(x) = \frac{\gamma_A(x)}{\gamma_B(x)} \frac{\gamma_B(x)}{\gamma_A(x)} x = x.$$
Proposition 12. $\gamma_B \circ \varphi_{A, B} = \gamma_A$.
Proof. Fix $x \in \Bbb{R}^n$. If $x = 0$, then clearly $\gamma_B(\varphi_{A, B}(x)) = \gamma_B(0) = 0 = \gamma_A(x)$. So, let us assume that $x \neq 0$, and so, by Proposition 1,
$$\gamma_B(\varphi_{A, B}(x)) = \gamma_B\left(\frac{\gamma_A(x)}{\gamma_B(x)}x\right) = \frac{\gamma_A(x)}{\gamma_B(x)}\gamma_B(x) = \gamma_A(x).$$
$\square$
Corollary 13. $\varphi_{A, B}(A) = B$ and $\varphi_{A, B}(\partial A) = \partial B$.
Proof. By Proposition 4, we have $A = \gamma_A^{-1}[0, 1]$ and $B = \gamma_B^{-1}[0, 1]$. By proposition 12, we have that
$$A = \gamma_A^{-1}[0, 1] = \varphi_{A, B}^{-1} \gamma_B^{-1}[0, 1] = \varphi_{A, B}^{-1}B,$$
but given proposition 11, $\varphi_{A, B}$ is invertible, hence $B = \varphi_{A, B}(A)$. A similar argument works to show $\varphi_{A, B}(\partial A) = \partial B$, simply by replacing $[0, 1]$ with $\{1\}$. $\square$
To summarise, if $A$ and $B$ are convex bodies in $\Bbb{R}^n$, the map $\varphi_{A, B}$ is a homeomorphism on $\Bbb{R}^n$ that maps $A$ to $B$, preserving boundaries. We now can answer your question.
Theorem 14. If $A$ and $B$ are convex bodies in $\Bbb{R}^n$, and $f : A \to B$ is continuous and maps $\partial A$ to $\partial B$ bijectively, then $f$ is surjective.
Proof. Without loss of generality, suppose $A, B$ contain $0$ in their interiors. Let $g : D^n \to D^n$ be defined by $g(x) = \varphi_{B, D^n}(f(\varphi_{D^n, A}(x)))$. By Proposition 11, $g$ is continuous. By Corollary 13,
$$g(D^n) = \varphi_{B, D^n}(f(A)) = \varphi_{D^n, B}(B) = D^n.$$
Similarly, by the same corollary, $g(\partial D^n) = \partial D^n$, and moreover, each step is a bijective map $\varphi_{D^n, A}$, $f$ and $\varphi_{B, D^n}$ are each bijectively mapping. Thus $g$ maps $\partial D^n$ to itself bijectively.
By Proposition 8, $g$ maps $D^n$ to itself surjectively. We can then use this prove $f$ is surjective.
Suppose $b \in B$. Let $y = \varphi_{B, D^n}(b)$. By Corollary 13, $y \in D^n$. Since $g$ is surjective, there exists some $x \in D^n$ such that $y = g(x)$. Note that:
$$\varphi_{B, D^n}(b) = y = \varphi_{B, D^n}(f(\varphi_{D^n, A}(x))) \implies b = f(\varphi_{D^n, A}(x)),$$
as $\varphi_{B, D^n}$ is injective. By Corollary 13 once more, $\varphi_{D^n, A}(x) \in A$, hence surjectivity is proven. $\square$
