If I have a monoid $\displaystyle S$, and an element $\displaystyle u\in S$, then the following equivalence holds:

$\displaystyle u$ is a unit iff $\displaystyle f(x)=xu$ and $\displaystyle g(x)=ux$ are bijections onto $\displaystyle S$.

Proof.Suppose $\displaystyle u$ is a unit. Let $\displaystyle x,y\in S.$ Then we have

1. $\displaystyle x=xu^{-1}u=f(xu^{-1});$

2. if $\displaystyle f(x)=f(y),$ then $\displaystyle xu=yu$ so $\displaystyle x=xuu^{-1}=yuu^{-1}=y.$

Conversely, if $\displaystyle f$ and $\displaystyle g$ are bijections onto $\displaystyle S,$ then there exists $\displaystyle x\in S$ such that $\displaystyle xu=f(x)=1$ and $\displaystyle y\in S$ such that $\displaystyle ux=g(x)=1.$ Thus $\displaystyle u$ has a left inverse and a right inverse, which is known to imply that $\displaystyle u$ is a unit.

Now let's say $\displaystyle S$ is a semigroup without an identity element. Can there be an element $\displaystyle u\in S$ such that $\displaystyle f(x)=xu$ and $\displaystyle g(x)=ux$ are bijections? Is it possible when $\displaystyle S$ is finite? (Then it's equivalent to asking whether a semigroup without ideantity can have a cancellable element.) Is it possible when $\displaystyle S$ is infinite? Is it possible when we only demand thatat least oneof $\displaystyle f,g$ be a bijection onto $\displaystyle S?$