Description of Green's relations $\mathcal{L}$, $\mathcal{R}$ in an inverse semigroup Theorem: Let $S$ be an inverse semigroup, and let $x,y\in S$ and $e,f\in E_{S}$ then


*

*$x\mathcal{L}y$ if and only if $x^{-1}x=y^{-1}y$

*$x\mathcal{R}y$ if and only if $xx^{-1}=yy^{-1},$
where $E_S$ denotes the set of idempotents of $S$, and $\mathcal{L}$, $\mathcal{R}$ are Green's relations.
For the forward of 1) I argue that if $x\mathcal{L}y$ then $L_{x}=L_{y}$. And since S is an inverse semigroup, S is regular and hence each $\mathcal {L}$-class contains a unique  idempotent. 
$x^{-1}x$ is idempotent in $L_{x}$ (why?), and $y^{-1}y$ is also idempotent in $L_{y}$ (why?). Hence $x^{-1}x=y^{-1}y$ Since each class contains a unique idempotent.
To be sincere, I did not convince myself about my proof, but I know somebody will surely point out my mistakes and corrected me.
Thanks.      
 A: Your proof is flawed, but not very much.
You say, "And since S is an inverse semigroup, S is regular and hence each $\mathscr{L}$-class contains a unique idempotent." 
No, it's not true that in a regular semigroup each $\mathscr{L}$-class contains a unique idempotent. The theorem is this:

Theorem 1. A semigroup $S$ is regular if and only if each $\mathscr L$- and $\mathscr R$-class contains an idempotent.

There's nothing about uniqueness here, and indeed those idempotents may not be unique. (Try finding an example!) However, there is another theorem:

Theorem 2. A semigroup $S$ is inverse if and only if each $\mathscr L$- and $\mathscr R$-class contains a unique idempotent.

So the uniqueness you're using follows from inverseness of $S$, not from regularity.
Now, you're asking why $x^{-1}x$ is an idempotent in $L_x$. But this is an easy exercise. We have $$(x^{-1}x)^2=(x^{-1}x)(x^{-1}x)=(x^{-1}xx^{-1})x=x^{-1}x.$$
This only uses the definition of the inverse. Now why is it in $L_x$? It is enough to show that there exist $a,b\in S^1$ such that $x=ax^{-1}x$ and $x^{-1}x=bx.$ Take $a=x$ and $b=x^{-1}.$
Now the other direction. Since $x^{-1}x$ and $y^{-1}y$ are equal and in the same $\mathscr L$-class as $x$ and $y,$ we have $$S^1x=S^1x^{-1}x=S^1y^{-1}y=S^1y$$ and so $x\mathscr L y.$
EDIT
I will give the proofs of theorems 1 and 2 here. First we will note the following fact.

Fact. Let $S$ be a regular semigroup and $x\in S.$ Then $S^1x=Sx$ and $xS^1=xS.$

I will leave the simple proof to you. This fact has the following corollary.

Corollary. Let $S$ be a regular semigroup and $x,y\in S.$ Then $$(1)\;\;x\mathscr L y \iff Sx=Sy$$ and $$(2)\;\;\;\;x\mathscr R y \iff xS=yS.$$

Proof of theorem 1. Suppose $S$ is a regular semigroup. Let $x\in S.$ Then there is $y$ such that $xyx=x$ and we have $(yx)^2=yx.$ We note that $x=x(yx)$ implies $Sx\subseteq S(yx)$ and $(yx)=y(x)$ implies $S(xy)\subseteq Sx.$ Therefore $$Sx=S(yx)$$ and $$x\mathscr L (yx).$$ Hence $L_x$ contains an idempotent, namely $yx$. We prove that the idempotent $xy$ is contained in $R_x$ similarly.
Conversely, suppose each $\mathscr L$- and $\mathscr R$-class contains an idempotent. Let $x\in S.$ We have idempotents $e^2=e\in L_x$ and $f^2=f\in R_x.$ There exist $a,b,c,d\in S^1$ such that $$x=ae,\;\;\;e=bx,\;\;\;x=fc,\;\;\;f=xd.$$ (Notice that we can't write $a,b,c,d\in S.$ We don't know $S$ is regular yet.)
Now we have $$x=ae=aee=xe=xbx=fcbx=ffcbx=fxbx=x(dxb)x,$$ where $b,d\in S^1.$ If $b,d\in S,$ then $dxb\in S$ and we are done. If $b\not\in S,$ then $b=1.$ Then if $d\in S,$ we have $dxb=dx\in S$ and we are done. But if also $d=1,$ then $dxb=x\in S$ and we are done. Similarly, nothing bad happens if $d=1.$ $\square$
Remark 1. Note that we could have been satisfied with $x=xbx.$ Indeed, then if $b\in S,$ the proof is done. If $b=1,$ then $x=xx=xxx$ and we are also done. I posted the previous calculation because the formula we obtain doesn't prefer either of the elements $b$ and $d.$
Remark 2. Note also that we have $$(bxd)x(bxd)=b(x(dxb)x)d=bxd.$$
Proof of theorem 2. Suppose $S$ is an inverse semigroup. It is therefore a regular semigroup and from theorem 1, we know that for every $x\in S$ we have idempotents $e^2=e\in L_x$ and $g^2=g\in R_x.$ We need to prove that they're unique. 
Suppose $f^2=f\in S$ and $e\mathscr L f.$ Then there exist $a,b\in S$ such that $$ae=f,\;\;\;bf=e.$$ But then $$efe=bffe=bfe=ee=e$$ and analogously $$fef=f.$$ Therefore $f$ is an inverse of $e.$ But also $e$ is an inverse of $e$ because $eee=e.$ Since in an inverse semigroup we have unique inverses, this gives us that $e=f.$ Analogously, we prove that idempotents in $\mathscr R$-classes are unique.
Conversely, suppose there are unique idempotents in $\mathscr L$- and $\mathscr R$-classes of $S.$ From theorem 1, we know that this implies the regularity of $S.$ This ensures the existence of inverses. We need to prove their uniqueness. Let $x\in S$ and let $x_1,x_2\in S$ be inverses of $x.$ It should be clear by now that it implies $$(x_1x)\mathscr L (x_2x),\;\;\;(xx_1)\mathscr R (xx_2).$$ But those elements are idempotents so we must have $$x_1x=x_2x,\;\;\;xx_1=xx_2.$$ Therefore $$x_1=x_1xx_1=x_2xx_1=x_2xx_2=x_2.$$ We proved that inverses are unique in $S,$ which ends the proof of the theorem. $\square$
