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.

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

Edit: there is a problem in what follows; the fact that $\beta$ must be contained in the interior of $\alpha$ is not true. It is possible that another normalization makes it hold, but I know feel that the postscriptum of the question is the good point of view.

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.

I think the following gives a proof, but I might have overlooked something.

Take two points $a,a'$ on $\alpha$, and let $b=f(a),b'=f(a')$. Apply an isometry of $\mathbb{R}^2$ to $\beta$, so as to get $a=b$ and $a'=b'$. Consider one of the segment from $a$ to $a'$ on $\alpha$ and name it $s$; then the points having same distance from $a$ and $a'$ than a point on $s$ form two arcs: $s$ itself and its mirror image with respect to $(aa')$. Due to convexity, one of the two segment between $a$ and $a'$ along $\beta$ must be equal to $s$. Proceed similarly for the other side.

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.