Let $M$ be a finite monoid and $e,e' \in M$ two distinct idempotent elements. Suppose we have some $x \in M$ such that $xe' \mathcal R e$ and $e \, \mathcal D \, e'$, where $\mathcal R$ and $\mathcal D$ denote Green's relations.
Let $v \in M$ be such that $xe'v = e$. Then as $ab \, \mathcal J\,a$ implies $ab\,\mathcal R\,a$ and $ab\,\mathcal J \, b$ implies $ab \, \mathcal L \, b$ and $\mathcal J = \mathcal D$ in a finite monoid we have as $xe'\,\mathcal R\, e$ also $xe' \,\mathcal D \,e$ and with $e \,\mathcal D \,e'$ also $xe' \,\mathcal D \,e'$, which gives $xe' \,\mathcal L \,e'$ and hence $xe'v \,\mathcal L \, e'v$. This also implies $e = xe'v\,\mathcal D \,e'v$ and again $e' \,\mathcal D \, e'v$, which gives $e' \, \mathcal R \, e'v$.
Taken together we have $$ xe' \, \mathcal R \, e = xe'v \, \mathcal L \, e'v \, \mathcal R \, e'. $$ Now in general we have for $m, m'' \in M$ $$ mm'' \in R_m \cap L_{m''} \mbox{ iff } R_{m''} \cap L_m \mbox{ contains an idempotent } $$ and as in our situtation $e \in R_{xe'} \cap L_{e'v}$ we have $(e'v)(xe') \in R_{e'v} \cap L_{xe'}$. Further as written above $e' \in R_{e'v}\cap L_{xe'} = H_{e'}$ we have $(e'v)(xe') \, \mathcal H \, e'$.
Now why is $(e'v)(xe')$ the product of two elements from the $H$-class $H_{e'}$ of $e'$?
I guess $e'v, xe' \in H_{e'}$ is not possible, for then with a similar reasoning as above we would have $(xe')(e'v) \in H_e$, giving $H_e = H_{e'}$ if $H_{e'}$ is a group (a fundamental result is that if $H^2 \cap H \ne \emptyset$ for a $H$-class it is a group), and thereby also closed under product, which would give $e = e'$ as every $H$-class has at most one idempotent element.
But I do not see that either $vxe' \in H_{e'}$ or $e'vx \in H_{e'}$? So does anybody see it?
Seems to be I was blind all the time, but now I see the simple implication that if a $H$-class contains an idempotent $e'$ then as $e'e' = e' \in H$ this gives that $H^2 \cap H \ne \emptyset$. This has nothing to do with the product $(e'v)(xe')$ and any decomposition thereof. Sorry...