Let $D$ be the closed disk, $B$ be the interior of $D$, $S^1$ the boundary.
I'll assume when you say $f:D\to D$ is continuous up to the boundary you just mean $f$ is continuous.
Suppose $f$ were not surjective. Then there is $a\in B\setminus f(D)$. Projecting outwards to the boundary, we get a map $D\to S^1$ which is a homeomorphism on the boundary. Composing with the inverse of $f|_{S^1}$, we get a map $r:D\to S^1$ such that $r|_{S^1}$ is the identity. I.e., we get a retraction of $D$ to $S^1$, which is impossible by the no retraction theorem.
As for injectivity, you say you know that $f^{-1}(x)$ has the same cardinality for all $x\in B$. We can do better. Observe that $f|_B:B\to B$ is a covering map (the restriction is well defined, since $f$ is a local homeomorphism on $B$, so interior points of $B$ can't map to boundary points of $D$). It's already a surjective local homeomorphism, and moreover $f^{-1}(x)$ for $x\in B$ is a closed subset of $D$, which is compact. Therefore $f^{-1}(x)$ is finite. Taking disjoint open neighborhoods $U_i$ of the points $p_1,\ldots,p_n \in f^{-1}(x)$ on which $f$ is a local diffeomorphism, we have $V = \bigcap_{i=1}^n f(U_i) \setminus f\left (D\setminus \bigcup_{i=1}^n U_i\right)$ is a trivializing open neighborhood of $x$, showing that $f$ is a covering map.
There are a couple of ways to see that this implies that $f|_B$ is injective. The easiest is that $B$ is simply connected, so the covering map $f:B\to B$ must be a universal cover for $B$, and the cardinality of the fibers of the universal cover is the same as the cardinality of the fundamental group, which is one in this case.
Another way to see this is that the identity map on $B$ has to lift over $f$. I.e., for any $p_i$ in the fiber of some point $x\in B$, there must be a map $g_i:B\to B$ such that $f\circ g_i = \operatorname{id}_B$ and $g_i(x)=p_i$. Restricting to some trivializing open set $V$ for $f$, we see that $g_i$ is a local homeomorphism, and therefore open. Thus $g_i(B)$ is an open subset of $B$, and every point of $B$ lies in some $g_j$ of $B$ (by uniqueness of lifts, we can't miss any points in any of the other fibers or hit them twice). Thus we have a partition of $B$ into a finite union of copies of $B$. Since $B$ is connected, we must have that there is only one copy of $B$. Hence the fibers of $f$ contain only one element.
Either way, we now know that $f$ is injective, since it sends $B$ to $B$ and $S^1\to S^1$ and is injective on each piece.
Thus as a continuous bijection from a compact space to a Hausdorff space, it must be a homeomorphism.
Edit I've fixed the proof that $f$ was a covering map. In the original version it wasn't clear that $f^{-1}(V)$ would be a disjoint union of open subsets of the $U_i$, but that was fixed by deleting the complement of the image of the rest of $D$ (the image being closed since $D$ is compact). The idea comes from this answer here. Also initially I was missing the word disjoint when defining the neighborhoods $U_i$, which is also fixed.
Second edit I'm fairly sure I've fixed my error proving that the map is a covering map, but here's a second argument following the suggestion of Moishe Kohan in the comments.
First, note that $f|_B$ is proper, since if $K$ is a compact subset of $B$, then it's also compact as a subset of $D$, so its preimage is closed in $D$, and thus compact, however the preimage must lie in $B$. Thus $f|_B$ is proper.
Then a proper, local homeomorphism is a covering map. See this answer here which generalizes the claim in the question to compactly generated spaces.
