I am not sure, some of my attempts so far had gave me trivial rings.
However I can illustrate how it works. For example consider the ring of integers $\mathbb{Z}$ and let's modify the distributive law so that + distribute over *, and left anything else unchanged. Then one can deduce that every integer will become idempotent under multiplication. Howeve for this example, eventually you get $1x=1$ thus the whole thing just collapse into the trivla ring. I have not explored further what axioms I can relax to avoid the collapse, though, because it is not my main focus atm

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…