Check $x=0$: $f(x)=|x-\pi|(e^{|x|}-1)\sin|x|$. Near $x=0$, both $(e^{|x|}-1)$ and $\sin|x|$ have factors of $|x|$, so $f(x)\approx |x-\pi|\cdot|x|^2$ near 0, which is differentiable at 0.

Check $x=\pi$: near $\pi$, $(e^{|x|}-1)\sin|x|$ is smooth and non-zero ($e^\pi\sin\pi=0$!). Since $\sin\pi=0$, the product $(e^{|x|}-1)\sin|x|$ vanishes at $\pi$, making $f(x)=0$ near $\pi$ — wait, $\sin\pi=0$, so $f(\pi)=0$. Checking differentiability more carefully shows $f$ is differentiable at $\pi$ too.

Therefore $S=\boldsymbol{\emptyset}$.