Let $a$ be an element of $A$. Then $a$ can be expressed as product of two elements, each of which can be expressed as a product of two elements, and so on forever. By the finiteness of $A$, among these expressions for $a$, some $y\in A$ appears to arbitrarily high powers.
Again by finiteness, we have $y^i=y^j$ for some positive integers $i
We conclude that there is an identity element $1_a$ for every $a\in A$.
Now apply repeatedly the following easy to verify lemma:
Lemma: If $1_u$ is an identity element for $u$, and $1_v$ is an identity element for $v$, then $1_u+1_v-1_u1_v$ is an identity element for both $u$ and $v$. (By $s-t$ we mean $s$ plus the additive inverse of $t$.)
Comment: An equivalent way to finish the argument is to let $M$ be a maximal subset of $A$ for which there is a $u\in A$ such that $um=m$ for all $m\in M$. If $M$ is not all of $A$, we can use the lemma to extend $M$.
Or else we can obtain an explicit and even symmetric expression for a unit in terms of all the $1_a$.