This problem is pretty well known.

I have to admit that my proof is not as ingenious as Tintarn's. Let the incircle touch the sides $AB,BC,CA$ in $U,V,W$. Then let \[ u = |WA| = |AU|, \qquad v = |UB| = |BV|, \qquad w = |VC| = |CW|. \]To prove $S \ne I$, proceed as Tintarn did. For the rest we'll just resort to some length bashing. Let $L$ be the intersection of the altitude onto $AB$ with $AB$. Lemma. We have $|CL| = 2 \sqrt{2uv}$. Proof. Compute the area of $\triangle ABC$ in two ways. (Heron and 1/2 times base times altitude) $\hfill \square$ Variation 1. Take $M_C$ as the midpoint of $AB$. By Pythagoras and $CS : SM_C = 2:1$ it is easy to compute $M_CA,M_CL,M_CS,M_CC$ respectively. We can conclude $\tfrac{M_CA}{M_CS}=\tfrac{M_CL}{M_CC}$. So there is a homothety with center $M_C$ that takes $\triangle UM_CS$ to $\triangle LM_CC$. That implies $SU \perp AB$, so we're done. (Watch out for different cases while length bashing.) $\hfill \square$ Variation 2. Coordinate bash! Use Cartesian coordinates and put $A = (0,0)$ and $B=(u+v,0)$ giving us $U=(u,0)$. Compute $AL^2=AC^2-CL^2=(2u-v)^2$ by Pythagoras. Now we can use the law of cosines to prove $x_C = 2u-v$, so \[ x_S = \frac{x_A+x_B+x_C}{3} = \frac{0+(u+v)+(2u-v)}{3}=u \]with which the result follows. $\hfill \square$