 Solution

For $a\in A$, the element $a=ae$ belongs to both $A$ and $aA$. The assumed equal-or-disjoint property therefore gives $aA=A$.

If $a,b\in A$, then $ab\in aA=A$, so $A$ is closed under multiplication. Also $e\in A=aA$, so some $c\in A$ satisfies $ac=e$. Therefore $c=a^{-1}\in A$. Along with the given identity element, closure and existence of inverse elements establish that $A$ is a subgroup of $G$. This is the subgroup test. The proof even works without assuming that $G$ is finite.