If a seimigroup's left identity is unique, can it be two-side identity? 
If a seimigroup's left identity is unique, can it be two-side identity?  

The answer is true if we talk it in a ring. Like construct $(be-b+e)a=a$. But in a semigroup I can't image how to construct a equation so we can use the condition uniqueness. So I am wondeting if the statement is false. But I also couldn't get a counterexample.

Consider a semigroup $S$ of words on letters $e, x$ of the form $x^n$ for $n>0 $ and $x^ne$ for $n\ge0$, defining
$$x^ne\cdot x^m:=x^{n+m}\quad\quad(n\ge0, m>0)\\
x^ne\cdot x^me:=x^{n+m}e\quad\qquad(n,m\ge0)$$
Here $e$ is the unique left identity, but it's not right identity as $xe\ne x$. 
Basically, $S$ is the quotient of the free semigroup of words on letters $e,x$, by the relations $ex\sim x,\ e^2\sim e$. 
