Let $T$ be foot of $I$ on $BC$. We see that inversion $(W;WI)$ maps $P\mapsto T$. And we have two results: a) $(SPIT)$ is fixed, $WI$ is tangent to it, $SI\parallel DE$. b) $T\in PW$. So it's a Simson line of $I$ and hence $I\in (DSE)$. With all this $DSIE$ - isoscales trapezoid.

Let $A'$, $K$ be the antipode of $A$ on $(ABC)$ and the foot of the perpendicular from $I$ to $\overline{BC}$, respectively. It's well-known that $P \equiv \overline{WK} \cap \overline{A'I}$. By the radical center theorem for $(ABC)$, $(AIP)$ and the incircle of $\triangle ABC$ we get that $\overline{SI}$ is tangent to $(BIC)$, and thus is perpendicular to $\overline{AW}$ and parallel to $\overline{DW}$. So, $\frac{SD}{AS}=\frac{WI}{AI}$, so it suffices to prove that $\frac{IE}{IW}=\frac{AS}{AI}$. For this notice the cyclic quadrilaterals $IKWE$ and $IKPS$, and so $\angle WIE=\angle WKE=\angle PKS=\angle PIS=\angle SAI$, so $\triangle SAI \sim \triangle EIW$, and the conclusion follows. $\blacksquare$