Let $e,f$ be idempotent of semigroup $S$ .
If $eLf$ then $ef=f$ and $fe=e$ .
From $eLf$ we have some $x,y \in S$ .
Such that $xe=f$ and $yf=e$.
Then $fxe = ff = f = xe $.
Can I reduce above to $ fx = x $ then $x=f$ .
If I can it will be clear.
Let $e,f$ be idempotent of semigroup $S$ .
If $eLf$ then $ef=f$ and $fe=e$ .
From $eLf$ we have some $x,y \in S$ .
Such that $xe=f$ and $yf=e$.
Then $fxe = ff = f = xe $.
Can I reduce above to $ fx = x $ then $x=f$ .
If I can it will be clear.
Your semigroup $S$ is not necessarily cancellative, so no, you can't reduce it.
To show $ef=f$ and $fe=e$ given $e\mathcal Lf$ for idempotents $e$ and $f$, you indeed have some $x, y\in S^1$ such that $xe=f$ & $yf=e$. Then $ef=(yf)f=yf=e$ and $fe=(xe)e=xe=f$.