If $ef = fe = e$ for two distinct idempotents in finite monoid, then $MeM \ne MfM$ Let $e,f$ be two distinct idempotent elements of some finite monoid such that 
$$
 ef = fe = e.
$$
Then of course $MeM \subset MfM$. But is $MeM = MfM$ possible? 

Do you know a proof that $MeM \ne MfM$, i.e. the two-sided ideals are different, or is $MeM = MfM$ possible?

I see that $f \notin eM \cup Me$, for assume $f = em$, then $e = ef = eem = em = f$, similar $f = me$ is not possible. But $f = men$ for $m,n \in M$ would not give any contradiction so far... 
 A: No, it is not possible (not even if you replace "monoid" by "semigroup").
See Theorem 3 below.
Let me first build up some (standard, I assume) arsenal.

Lemma 1. Let $M$ be a finite monoid with neutral element $1$. Let $x$ and $y$ be two elements of $M$ such that $xy=1$. Then, $yx=1$.

Proof of Lemma 1. Let $\alpha$ be the map $M\rightarrow M,\ m\mapsto mx$.
Let $\beta$ be the map $M\rightarrow M,\ m\mapsto my$. Then, every $m\in M$ satisfies
$\left(  \beta\circ\alpha\right)  \left(  m\right)  =\beta\left(
\underbrace{\alpha\left(  m\right)  }_{\substack{=mx\\\text{(by the definition
of }\alpha\text{)}}}\right)  =\beta\left(  mx\right)  =m\underbrace{xy}_{=1}$
(by the definition of $\beta$)
$=m=\operatorname*{id}\left(  m\right)  $.
In other words, $\beta\circ\alpha=\operatorname*{id}$. Hence, the map $\beta$
is right-invertible, therefore surjective. But the set $M$ is finite. Hence,
every surjective map $M\rightarrow M$ is bijective. Applying this to the
surjective map $\beta:M\rightarrow M$, we conclude that the map $\beta$ is
bijective. Hence, the map $\beta$ is invertible. The inverse $\beta^{-1}$ of
this map $\beta$ must be $\alpha$ (since $\beta\circ\alpha=\operatorname*{id}
$). Thus, $\alpha\circ\beta=\operatorname*{id}$. Now, the definition of
$\alpha$ yields $\alpha\left(  y\right)  =yx$. The definition of $\beta$
yields $\beta\left(  1\right)  =1y=y$. But
$\left(  \alpha\circ\beta\right)  \left(  1\right)  =\alpha\left(
\underbrace{\beta\left(  1\right)  }_{=y}\right)  =\alpha\left(  y\right)
=yx$,
so that $yx=\underbrace{\left(  \alpha\circ\beta\right)  }
_{=\operatorname*{id}}\left(  1\right)  =\operatorname*{id}\left(  1\right)
=1$. This proves Lemma 1. $\blacksquare$

Lemma 2. Let $M$ be a finite monoid with neutral element $1$. Let $e$ be an idempotent element of $M$ such that $1\in MeM$. Then, $e=1$.

Proof of Lemma 2. We have $1\in MeM$. In other words, there exist $u\in M$
and $v\in M$ such that $1=uev$. Consider these $u$ and $v$.
We have $uev=1$. Thus, Lemma 1 (applied to $x=u$ and $y=ev$) yields $evu=1$.
But $e$ is idempotent, so that $e^{2}=e$. Now, $\underbrace{e^{2}}
_{=e}vu=evu=1$, so that $1=e^{2}vu=e\underbrace{evu}_{=1}=e$. This proves
Lemma 2. $\blacksquare$

Theorem 3. Let $M$ be a finite semigroup. Let $e$ and $f$ be two idempotent elements of $M$ such that $MeM=MfM$ and $ef=fe=e$. Then, $e=f$.

Proof of Theorem 3. Since $f$ is idempotent, we have $f=f\underbrace{f}
_{=ff}=fff$. Let $M^{\prime}$ be the subsemigroup $fMf$ of $M$. Then,
$f = f\underbrace{f}_{\in M}f \in fMf = M^\prime$.
Moreover, the element $f$ of $M^\prime$ is a neutral element with
respect to multiplication (since
each $m \in M$ satisfies $f \left(fmf\right) = \underbrace{ff}_{=f} mf
= fmf$ and similarly $\left(fmf\right) f = fmf$).
Hence, the subsemigroup $M^{\prime}$ of $M$
is a monoid with neutral element $f$.
We have $e=f\underbrace{e}_{=ef}=f\underbrace{e}_{\in M}f\in fMf=M^{\prime}$.
We have $f=\underbrace{f}_{\in M}f\underbrace{f}_{\in M}\in MfM=MeM$. But
$f=f\underbrace{f}_{\in MeM}f\in fM\underbrace{e}_{=fef}Mf =
\underbrace{fMf}_{=M^{\prime}}e\underbrace{fMf}_{=M^{\prime}}=M^{\prime
}eM^{\prime}$. Hence, Lemma 2 (applied to $M^{\prime}$, $f$ and $e$ instead of
$M$, $1$ and $e$) yields that $e=f$. This proves Theorem 3. $\blacksquare$
