It seems there should a proof along the following lines, but I did not took the time to check every detail.

First, the assumptions give that $f$ is an isometry with respect to the length metrics on $\alpha$ and $\beta$. Let $a$, $a'$ be two farthest points on $\alpha$, $b=f(a)$ and $b'=f(a')$. Without lost of generality, apply to $\beta$ an isometry of $\mathbb{R}^2$ so that $b=a$ and $\beta$ is contained in the half-plane delimited by the line orthogonal to $[aa')$ at $a$. Using that $f$ is $1$-Lipschitz and that $[aa']$ is a diameter, we get that $\beta$ must be contained in the interior of $\alpha$. Considering the projection to the domain delimited by $\beta$, we get a $1$-Lipschitz map $\tilde f:\alpha\to\beta$ that contracts strictly distances around any point $x\in\alpha$ such that $\alpha$ at $x$ and $\beta$ at $f(x)$ do not share a common supporting direction. Since $\alpha$ and $\beta$ have the same length, this never happens and $\beta$ must be an homothetic image of $\alpha$. Since they have the same length, the homothety constant must be $1$ and we are (hopefully) done.