Romanian Mathematical Olympiad

If $1+1\ne 0$, then $x\ne -x$, for every invertible $x$, whence $S=0$ (we can group the invertible elements into pairs $(x,-x)$).

If this is not the case, notice that $xS=S$, for every invertible $x$. Adding all these relations, $S^2=kS$, where $k$ is the number of invertible elements. Then $S^2=S$ for odd $k$ and $S^2=0$ for even $k$ (since $1+1=0$).