Proposition: Given a semigroup $S$. If $\exists a,b \in S, ab=1$ and $\forall x\in S, 1x=x$ and $bx\neq by, \forall x,y \in S$, then $a: x \to ax$ is bijective and hence a permutation.
Proof: Since $bx\neq by, \forall x,y$, $\exists z \in S, bx=z, \forall x$ (i.e. $b: x\to bx$ is injective. Next multiply by $a$ on the left to get $a(bx)=az$. By associativity $a(bx)=(ab)x$ and by $ab=1$, $abx=1x$ and then by $1x=x,\forall x$, $1x=x$. Since $1$ is a left identity, it is injective wrt all $x$. Hence $a(bx)=1x=x$. Since $1$ is injective and $bx\in S$ is unique due to injectivity of $b$, $a$ is…