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 $i<j$. Take a representation of $a$ as a product that uses a power of $y$ which is $i$ or greater. Then $i$ of these $y$'s can be replaced by $j$ $y$'s. This procedure does not change $a$, but it multiplies $a$ by $y^{j-i}$. Thus $a=y^{j-i}a$, and therefore there is an identity element for $a$.

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 $1_a$.