Let $a,b$ be elements of a regular semigroup $S$. Then $(a,b) \in \mathcal L$ iff there exist an inverse $a'$ of $a$ and $b'$ of $b$ such that $a' a = b'b$.
I have done the converse parts: if there exist an inverse $a'$ of $a$ and $b'$ of $b$ such that $a' a = b'b$ , then $(a,b) \in \mathcal L$.
Suppose $(a,b) \in \mathcal L$, so $a,b$ belongs the same $\mathcal D-$ class says $D$ and $a,b$ are regular, so every $\mathcal R$ class in $D$ contains an idempotent , it follows that $\mathcal R_b$ contains an idempotent says $e$. how to proceed further .
Any help would be appreciated, Thank you
Suppose that $(a,b) \in \mathcal{L}$. Then there exists an invese $a'$ of $a$ such that $a'a$ is an idempotent in $L_a = L_b$. You know that every $\mathcal{R}$ contains an idempotent says $e$, which implies that $\exists \ b' \in R_{a'a} \cap L_e$ such that $b'b=a'a$. The reason of $\exists \ b' \in R_{a'a} \cap L_e$ is explained as follows.
Suppose that $e \in R_b$. We get $(e, b) \in \mathcal{R}$, and $eb=b$, $bx=e$ for $x \in S^1$, where $S^1$ is obtained by adding $1$ to $S$ if necessary. Moreover, $(a'a,b) \in \mathcal{L}$ implies that $ba'a = b$, $yb = a'a$ for $y \in S^1$.
Take $b' = a'axe$. You could get $bb'b=b$, $b'bb'=b'$, $bb'=e$, $b'b=a'a$. So $b' \in R_{a'a} \cap L_e$.