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 $1 + 1 = 0$).