Let $S$ be a finite semigroup and let $e, f$ be idempotents of $S$. I want to show that $SeS = SfS$ is equivalent to the existence of $x, y \in S$ such that $xy = e$ and $yx = f$.
The second direction is pretty straightforward since $SeeS = SxyxyS \subset SyxS = SfS$. The other direction I know is true, it can be found in lists of equivalent statements in some textbooks, but I'm hoping for there to be a very nice direct proof.
Theorem. Two idempotents of a semigroup are conjugate if and only if they are $\cal D$-equivalent.
This result is proved for instance in [1, Proposition 2.3.5]. The proof relies on Green's Lemma, that I will not recall here. Let $R(x)$, $L(x)$ and $H(x)$ denote the $\cal R$-class, the $\cal L$-class and the $\cal H$-class of an element $x$, respectively.
Proof. Suppose that $e = uv$ and $f = vu$ for some $u, v \in S$. Then $uvuv = uv$ and $vuvu = vu$, whence $uv \mathop{\cal R} uvu$ and $uvu \mathop{\cal L} vu$. Thus $e = uv \mathop{\cal D} vu = f$.
Suppose now that $e \mathop{\cal D} f$. Then there exists $s \in S$ such that $e \mathop{\cal R} s$ and $s \mathop{\cal L} f$. By Green's lemma, there exists an element $\bar s \in L(e) \cap R(f)$ such that $\bar ss = f$. Thus $s\bar ss = sf = s$ and $\bar ss\bar s = f\bar s = \bar s$. It follows that $\bar s$ is an inverse of $s$. Thus $s\bar s$ is an idempotent of $H(e)$ and thus is equal to $e$.
Your question considers the condition $SeS = SfS$, which means that $e \mathop{\cal J} f$. It is a well-known fact that ${\cal J} = {\cal D}$ in a finite semigroup. This property also holds if, for every $\cal J$-class $J$ of $S$, the set of $\cal L$-classes (respectively $\cal R$-classes) contained in $J$ contains a minimal member. In particular, it holds in any periodic semigroup. It also holds in compact topological semigroups.
[1] J.~M. Howie, Fundamentals of semigroup theory, London Mathematical Society Monographs. New Series vol.~12, The Clarendon Press Oxford University Press, New York, 1995.