Here is an attempt at a constructive answer. The claim is that a pure sinusoid is the only periodic waveform that has the property $\forall A_1,\tau_1 \exists A_2,\tau_2$ such that $x(t) + A_1x(t+\tau_1) = A_2 x(t+\tau_2)\;\forall t$ where $x(t)$ is $T$-periodic.
I am going to limit myselfto "well-behaved" functions which can be uniquely expressed as a Fourier series, i.e., $$x(t) = \sum_\ell \gamma_\ell {\rm e}^{\jmath \ell t/T}.$$ The pure sinusoids are the functions where only one pair of $\gamma_{\ell_0}$, $\gamma_{-\ell_0}$ is nonzero and all other $\gamma_\ell$ are zero. For instance $\gamma_1 = \gamma_{-1} = \frac 12$ and all others zero gives $x(t) = \cos(2\pi t/T)$ but it could also be $\gamma_2$ and $\gamma_{-2}$, giving rise to a cosine of twice the frequency (whose fundamental period is then $T/2$, but it is still also $T$-periodic).
Now apply this expansion into the definition. We obtain $$\sum_\ell \gamma_\ell {\rm e}^{\jmath \ell t/T} + A_1 \sum_\ell \gamma_\ell {\rm e}^{\jmath \ell t/T} {\rm e}^{\jmath \ell \tau_1/T} \stackrel{!}{=} A_2\sum_\ell \gamma_\ell {\rm e}^{\jmath \ell t/T}{\rm e}^{\jmath \ell \tau_2/T}, $$ which we can write as $$\sum_\ell \gamma_\ell\cdot\left(1+{\rm e}^{\jmath \ell \tau_1/T}A_1\right) {\rm e}^{\jmath \ell t/T}\stackrel{!}{=}
\sum_\ell \gamma_\ell\cdot\left({\rm e}^{\jmath \ell \tau_2/T}A_2\right) {\rm e}^{\jmath \ell t/T}.$$
Now we can claim that for both sides of the equation to be equal for all $t$, all coefficients must be equal since each deviation in coefficients (for the same $\ell$) gives rise to a nonzero difference function and different coefficients (for different $\ell$) cannot cancel as the basis functions of the Fourier series are orthogonal. Therefore, the above condition translates to $$ \gamma_\ell\cdot\left(1+{\rm e}^{\jmath \ell \tau_1/T}A_1\right) \stackrel{!}{=}
\gamma_\ell\cdot\left({\rm e}^{\jmath \ell \tau_2/T}A_2\right) \; \forall \ell. $$
So again, given an arbitrary $A_1$ and $\tau_1$ we must find an $A_2, \tau_2$ such that the above condition is true for all $\ell$. There are two ways to satisfy the equation: either $\gamma_\ell = 0$ or $|A_2| = \sqrt{1+A_1^2 + 2A_1\cos(\ell \tau_1/T)}$ (using $|1+A|=\sqrt{((1+\Re A)^2 + (\Im A)^2}$) and $\tau_2= \frac{T}{\ell} \arg\left\{\frac{1+{\rm e}^{\jmath \ell \tau_1/T}A_1}{A_2}\right\}$. Now, obviously the solution for $A_2$ will be different for each $\ell$ as long as $A_1 \neq 0$ (and for $\tau_1/\pi$ irrational). So we can solve $A_2$ only for one $\ell$ (and since the cosine is even, the same solution works for $-\ell$).
In consequence, we can have one pair $(\ell,-\ell)$ for which $\gamma_\ell \neq 0$ that allows us to solve for $(A_2, \tau_2)$ for any given $(A_1,\tau_1)$. However, since the same solution will not work for any other $\ell$, all other $\gamma_\ell$ need to be zero. Hence, the only solutions that work are of the form $$x(t) = \gamma_1 \cdot {\rm e}^{\jmath \ell_0 t/T} + \gamma_{-1}{\rm e}^{-\jmath \ell_0 t/T} = (\gamma_1 + \gamma_{-1}) \cos( \ell_0 t/T) + \jmath (\gamma_1 - \gamma_{-1}) \sin( \ell_0 t/T),$$
which are exactly the pure sinusoids of (radial) frequency $\ell_0/T$.
