Term for a semigroup with left identities and left inverses? Is there a term for a semigroup $(M, *)$ that has at least one
left identity and left inverses in the "weak" sense that, for all
$a \in M$, there exists a $b \in M$ such that $b*a$ is a left
identity of $M$? In other words, I'm looking for a term that assumes only the
following axioms:

*

*$M$ is nonempty.

*$* \colon M \times M \to M$.

*$*$ is associative.

*$\forall a \in M,\, \exists b \in M \text{ such that, } \forall
   x \in M,\, (b*a)*x = x$.

I know that such a semigroup is just a group if all elements have
a left inverse for the same left identity $e$, meaning
that, for all $a \in M$, there exists a $b \in M$ such that $b*a =
e$.  However, I'm looking for a term that does not assume that the
left inverses are all for the same left identity.
An example of such a semigroup that is not a group is the set
$$\left\lbrace \begin{bmatrix} x & y \\ x & y \end{bmatrix} : x, y \in \mathbb{R}, x + y \ne 0\right\rbrace$$
of $2 \times 2$ matrices under matrix multiplication, in which every element of the form $\begin{bmatrix} a & 1-a \\ a & 1-a \end{bmatrix}$ is a left identity, and, given $A = \begin{bmatrix} a & b \\ a & b \end{bmatrix}$, every element of the form $\begin{bmatrix} c & d \\ c & d \end{bmatrix}$ with $c + d = (a + b)^{-1}$ is a left inverse of $A$.
 A: These semigroups are called right groups. They have a very simple description. Let $S$ be a semigroup satisfying your condition.

*

*Every semigroup of right zeroes (that is a set with operation $*$ such that $\forall x, y: x*y=y$) satisfies this condition. The direct product of a group and a right zero semigroup satisfies this condition as well. I prove below that there are no other examples.


*Every idempotent in $S$ is a left identity. Indeed if $a^2=a, ba=e, \forall x:\quad  ex=x$, then $a=ea=baa=ba=e$.


*The set $E$ of all idempotents in $S$ is a right zero semigroup.


*The semigroup $S$ is ideally simple: every ideal is the whole semigroup. Indeed, if $I$ is an ideal, $a\in I$, and $ba=e$ is an idempotent, then  $e\in I, eS=S\subseteq I$, hence $I=S$.


*All idempotents in $S$ are primitive: if $ef=fe=f$, then $f=e$ (from 3.)


*Hence $S$ is a completely simple semigroup, so it is a Rees-Sushkevich semigroup $M(G;I,J,P)$. If $P$ has more than one row (i.e. $|J|>1$) , then the idempotents of $S$ do not form a right zero semigroup. Thus $|J|=1$ and $S$ is isomorphic to the direct product of the group $G$ and a right zero semigroup.
