This is exercise 11.19 (h) in Seth Warner's "Modern Algebra".
We are given that $S$ is an idempotent semigroup, that is: $\forall a \in S: aa = a$.
Let $R$ be the equivalence relation defined as: $$aRb \iff (aba=a) \land (bab=b)$$
We have that $R$ is a congruence relation on $S$, that is: $$aRb \land cRd \iff (ac) R (bd)$$
We have to prove that the quotient structure $S/R$ is a commutative idempotent semigroup.
Proving idempotence is trivial, as $[a]_R [a]_R = [aa]_R = [a]_R$.
But I can't work out where to start on commutativity.
I see $[a]_R [b]_R = [axabyb]_R$ where $x$ and $y$ are arbitrary such that $x R a$ and $y R b$ but I can't drive off from there.
During the course of the exposition to demonstrate that $R$ is a congruence relation, which is part (g) of this question, various properties about $S$ and $R$ are demonstrated, all building from things like "if $xy = y$ and $yx = x$ then $\forall z: zxzy=zy$ and $zyzx=zx$" but I can't see how to use that approach here. $axabyb$ is general, and there is no suggestion that any of those properties apply there.
For context, I am self-taught and I am doggedly plugging away at this work question by question.
It suffices to prove that $ab \mathrel{R} ba$, that is, according to the definition of $R$, $(ab)(ba)(ab) = ab$ and $(ba)(ab)(ba) = ba$. But since $S$ is idempotent, one has $$(ab)(ba)(ab) = ab^2a^2b = abab = (ab)^2 =ab.$$ The second equality is dual (just exchange $a$ and $b$).