I proved this in the following way is our exam, but got $0$ out of $7$ points for it, however I fail to see why its wrong:
Let $a\in S$ be arbitrary. Since $S$ is finite, there has to be some $m \in \mathbb{N}$ such that $a^m=e$, otherwise all the powers of $a$ would be different. Then $a^m = a^m(a^{-1}a)$, and by cancellation we get $e = a^{-1}a$, so that $a^{-1}$ is the left inverse.
Similarly I showed a right inverse exists, hence a unique two sided inverse exists, so $S$ is a group. Being periodic, I know some power of $a$ has to be an idempotent, so that power becomes the identity.
I suppose that $e$ is the identity of your monoid. Why does there have to be a power of $a$ equal to the identity? As the commenter above says there have to be two powers of $a$ that are equal but not necessarily equal to the identity, at least without first proving you the monoid is a group. Also when you write $a^m = a^m aa ^{-1}$ what is $a ^ {-1}$? You'd maybe like to say the inverse of $a$, but we don't know that there is one yet. You seem to be using the fact that your monoid is a group to prove that it is a group. You see the problem with that, right?