This is an exercise from "Abstract Algebra" by P.A.Grillet (p.12, ex.2).
Let $S$ be a semigroup (that is, a set with an associative binary operation) in which there is a left identity element ($\exists e \in S: \forall x \in S \ \ ex = x$) relative to each every element of $S$ has a left inverse ($\forall x \in S \ \ \exists y \in S$ such that $yx = e$). Prove that $S$ is group.
The following proof, interestingly enough, essentially comes from Section 1.1 of EBBINGHAUS-FLUM-THOMAS Mathematical Logic:
First, fix a left identity $e$ (which is not unique, in principle). Let's show that for each $x\in S$ there exists $y\in S$ with $xy=e$. More precisely, any left inverse of $x$ is a right inverse of $x$.
Given $x$, choose $y$ for which $yx=e$, and then choose $z$ for which $zy=e$. Then \begin{align*} xy=e(xy)=(zy)(xy)=z((yx)y)=z(ey)=zy=e \end{align*}
Now let's show that the left unit $e$ is also a right unit, and we are done (this is not in the book cited above): Given $x\in S$, choose $y$ for which $yx=xy=e$, so $$xe=x(yx)=(xy)x=ex=x$$
So $S$ has a bilateral unit, and each element of $S$ has a bilateral inverse (with respect to this unit, which is unique, etc...), so $S$ is a group.