You do not need differentiability. Below is a proof that works when $\alpha$ is only continuous.
We may assume that $I$ starts with $0$ : $I=[0,M]$ for some $M$. Suppose that the image of $\alpha$ is not contained in a line. Then there are $x_1 such that $\alpha(x_1),\alpha(x_2),\alpha(x_3)$ are not collinear. Let $t\in ]0,M-x_3[$. Let now $x=x_4$ be an arbitrary number in $[0,M-t]$. Consider the two families of four points $A_i=\alpha(x_i) (1 \leq i \leq 4)$ and $B_i=\alpha(x_i+t) (1 \leq i \leq 4)$. They are isometric by hypothesis : $d(A_i,A_j)=d(B_i,B_j)$.
By this older StackOverflow question, there is an isometry $\gamma(x,t)$ of the plane sending $A_i$ to $B_i$. Since $\gamma(t)$ is already uniquely determined by its image on $\alpha(x_1),\alpha(x_2),\alpha(x_3)$, we see that $\gamma(x,t)=\gamma(t)$ is in fact independent of $x$, and we can explicitly write out the coefficients of the matrix of $\gamma(t)$ in terms of the coordinates of $\alpha(x_1),\alpha(x_2),\alpha(x_3)$ and $\alpha(x_1+t),\alpha(x_2+t),\alpha(x_3+t)$. So $\gamma$ will be continuous if $\alpha$ is. Unicity also ensures that $\gamma$ is a homomorphism (where it is defined) : $\gamma(s+t)=\gamma(s)\gamma(t)$ when $s,t$ and $s+t$ are all in $I$.
Now ${\sf det}(\gamma(t))=1$ or $-1$, and by continuity this determinant is always $1$ or always $-1$. Since $\gamma(0)$ is the identity, we see that $\gamma$ is always an (affine) rotation. We have $\gamma(s)\gamma(t)=\gamma(s+t)=\gamma(t)\gamma(s)$. We see that $\gamma(s)$ and $\gamma(t)$ commute, so they must share the same center point. We deduce that the center $\Omega$ of $\gamma(t)$ is independent of $t$. Now for any $s, $\alpha(t)=\gamma(t-s)\alpha(s)$, so $\alpha(t)$ and $\alpha(s)$ are located at the same distance from $\Omega$. So $\alpha$ walks on a circle of cnter $\Omega$, qed.