$eSe$ in a finite semigroup

Question

Where is the following argument going wrong?

Let $S$ be a finite semigroup. There exists $e\in S$ such that $ee=e$. The subsemigroup $eSe = \{ese \mid s\in S\}\subseteq S$ is a monoid with the identity $e$. The map $ese\mapsto s$ is an injection from $eSe \to S$. Therefore $eSe = S$. Thus, every finite semigroup is a monoid ?! What?!

Answer 1

The reason is, formula $ese\mapsto s$ does not always define a function, as there might be two different $s$ and $s'$ such that $ese = es'e$.