A combinatoric theorem about semigroups

Question

I am struggling with the following theorem about semigroups, so I was hoping someone could give me a hand.

The theorem states: "Let $S$ be an arbitrary semigroup such that for every $a\in{}S$ it holds $a\in aS^2$ and for every $a,x\in S$ the condition $a^2x=a$ implies that $x^2a=x$. Then for every $a,b\in S$ we have that $a\in abSa$"

I will show you my attempt. So far, I was only able to prvef that $S$ is completely regular. Indeed, consider any element $a\in S$. Then $a=a^2x$ for some $x\in S$, which implies that $x=x^2a$. But then \begin{align*} a&=a^2x=a^2x^2a=(a^2x)xa=axa,\\ x&=x^2a=x^2a^2x=(x^2a)ax=xax \end{align*} So element $a$ is regular (and therefore $S$ is a regular group) and if $a=a^2x$, then $x$ is the inverse of $a$. Now consider the element $a^2$. From the hypothesis, we know, that $a^2=a^4y$ which implies $y=y^2a^2$. Also from the previous, we know, that $y$ is inverse of $a^2$, so we get $$a=a^2x=(a^2ya^2)x=a^2y(a^2x)=a^2ya=a^2(y^2a^2)a=(a^2y^2a)a^2$$ This means that $a\in Sa^2$. So, now we have $a\in a^2S\land a\in Sa^2$ which implies, that $a$ is a completely regular element, therefore $S$ is a completely regular semigroup.

I don't know how to continue from here, so I would appreciate any help. Thanks