Ok. So it's quite elementary actually.

Let me see if I got this. Let $\widetilde{X}\longrightarrow \mathbf{P}^1_S$ be the normalization. Then, by the universal property of normalization, there is a unique morphism $q:X\longrightarrow \widetilde{X}$. Since $q$ is birational and quasi-finite, Zariski's main theorem implies that it is an open immersion. Now, since $f$ was finite, we have that $q$ is finite. Therefore, it is closed. Therefore, $q$ is the identity morphism.