Do these very VERY weak axioms guarantee a group? (Every element has left identity $e_L$ or right identity $e_R$, with same-sided inverse)

Question

I can't help but ask, after we've come so far weakening the group axioms in these two posts, whether we can get even weaker?

Let $A$ be a set with an associative binary operation $*$, and suppose there exist two elements $e_L,e_R\in A$ such that, for all $x\in A$, at least one of the following two conditions holds:

1. $e_L*x=x$ and there exists an $x'\in A$ such that $x'*x=e_L$;
2. $x*e_R=x$ and there exists an $x'\in A$ such that $x*x'=e_R$.

Must $(A,*)$ be a group?

The second linked post proves that the answer is “yes” in the case where $e_L=e_R$. This generalized version was posed by @Yakk in the comments to @Vincent's answer in the first linked post.

Answer 1

Take $A = \{1, 0\}$ with the usual multiplication. Let $e_L = 1$ and $e_R = 0$. If $x = 1$, then $e_Lx = x$ and $x'x = e_L$ for $x' = 1$. If $x = 0$, then $xe_R = x$ and $xx' = e_R$ for $x' = 0$.

Answer 2

If I'm not mistaken. Here's a counterexample. Let $G$ and $H$ be arbitrary groups, $A=G\cup H$. Define multiplication on the set $A$ by the rule

$$ xy= \left\{% \begin{array}{ll} xy, & \hbox{if $x,y\in G$ or $x,y\in H$;} \\ y, & \hbox{if $x\in G$ and $y\in H$;} \\ x, & \hbox{if $x\in H$ and $y\in G$.} \\ \end{array}% \right. $$ Here $e_L=e_G$, $e_R=e_H$.