A semigroup $S$ is $\mathcal{R}$-unipotent if its regular and $efe=ef$ for any idempotents $e,f$ in S.
I want to show that if $S$ is regular then $S$ is $\mathcal{R}$-unipotent if and only if its idempotents form a left normal band, that is, $gfe=gef$ for any idempotents in $e,f,g$ of $S$.
If $S$ is regular and its idempotents form a left normal band then, for any $e,f,g$ idempotents of $S$ we have
$$gfe=gef.$$ In particular, $$efe=eef=ef. $$
Reciprocally , let's assume $S$ is $\mathcal{R}$-unipotent. Let $e,f,g$ be idempotents of $S$. Then $$gfe=gfef=gf(gef) $$ and $$gef=gefe=ge(gfe). $$ Therefore $gfe \mathrel{\mathcal{L}}gef$. If I managed to show $gfe \mathrel{\mathcal{R}}gef$ since in a $\mathcal{R}$-unipotent semigroup each $\mathcal{R}$-class only has one idempotent I could conclude $gef=gfe.$ Any hints?
Counterexample. Take the 5-element idempotent monoid $M = \{1, a, b, ab, ba\}$ presented by $aba = ab$ and $bab = ba$. Then $efe = ef$ for all $e, f \in M$, but the identity $gef = gfe$ fails if you take $g = 1$, $e = a$ and $f = b$.