$\newcommand{\unsim}{\mathord{\sim}}$Let $G$ be a group. What is $$ G/\left(ab\sim ba\ \middle|\ a,b\in G\right)? $$ Answer: not $G^{\mathrm{ab}}$, but the set of conjugacy classes of $G$.
When passing to monoids, the situation gets more complicated: the equivalence relations generated by the following declarations are all equivalent for $M$ a group, but not for $M$ a monoid:
(1) Conjugacy. $a\sim_{1}b$ iff there exists an invertible $g\in M$ such that $gag^{-1}=b$;
(2) Conjugacy, II. $a\sim_{2}b$ iff there exists a non-necessarily invertible $m\in M$ such that $ma=bm$;
(3) "Commutativity". $ab\sim_{3}ba$ for each $a,b\in M$;
In an MO answer, Tom Leinster explained how to describe $M/\unsim_{1}$ and $M/\unsim_{2}$: First, let $[\mathbb{N},M]$ be the category where
- $\mathrm{Obj}([\mathbb{N},M])=M$;
- A morphism $m\longrightarrow m'$ is an element $g$ of $M$ such that $gm=m'g$.
Then we have bijections $$ \begin{align*} M/\unsim_{1} &\cong [\mathbb{N},M]/\{\text{isos}\},\\ M/\unsim_{2} &\cong \pi_{0}([\mathbb{N},M]). \end{align*} $$
Question: Given a monoid $M$, what is the set $M/\unsim_{3}=M/\left(ab\sim ba\ \middle|\ a,b\in M\right)$?