Product of non-idempotent elements in a commutative monoid

Question

Assuming we have a commutative monoid $(M,\cdot)$ such that the non-trivial elements have no inverse. In addition, M contains no non-trivial idempotents. Considering two non-trivial elements $a_1$ and $a_2$. Is it possible to have a relation of the form $a_{1}^{n_1}a_{2}^{n_2} = a_{1}^{n_1^{'}}a_{2}^{n_2^{'}}$ with $(n_1,n_2) \neq (n_{1}^{'},n_{2}^{'})$?
I am suspecting the answer is a no. But so far I've only been able to show that in the case where $n_1 > n_1^{'}$ and $n_2 > n_2^{'}$ we have an idempotent element $a_1^ta_2^m$ for some $t,m$. I have not been able to prove this when $n_1 > n_1^{'}$ and $n_2 \leq n_2{'}$.

Answer 1

Take the monoid $M = \langle a, b \mid ab = ba, a^2b^2 = ab^3\rangle$. Its elements are of the form $a^n$ ($n \geq 0$), $b^m$ ($m > 0$), $ab^m$ ($m > 0$) or $a^2b$ and the unique idempotent is $1 = a^0$.