I have a semigroup $S$ with the following property:
For $a,x\in S$, there exist a unique $x'\in S$ such that $$ axx' = a$$
Currently, I'm calling $x'$ the $a$-inverse of $x$. Is there a more standard notion? I'm interested known what is known for semigroups with this property.
The semigroups satisfying your property are exactly the right groups, which are semigroups of the form $G \times R$, where $G$ is a group and $R$ is a right zero semigroup. Here is a proof, which can probably be simplified.
Step 1: $S$ is regular. Note that $axx' = a$ implies $ax(x'xx') = a(xx')(xx') = a(xx') = a$. Thus, by unicity of the $a$-inverse of $x$, one obtains $x' = x'xx'$. Furthermore, setting $b = ax$, one gets $bx'x = axx'x = ax = b$ and $bx'(xx'x) = b(x'x)(x'x) = b(x'x) = b$. By unicity of the $b$-inverse of $x'$, one gets $x = xx'x$.
The two relations $x = xx'x$ and $x' = x'xx'$ show that $x'$ is necessarily an inverse of $x$ in the semigroup sense. Thus every element of $S$ has an inverse and thus $S$ is a regular semigroup.
Step 2: all the idempotents of $S$ are $\mathcal{R}$-equivalent. Let $e$ and $f$, let $x$ be an inverse of $ef$ and let $h = fxe$. Then $h$ is idempotent since $h^2 = (fxe)(fxe) = f(xefx)e = fxe$. Moreover $fh = ffxe = fxe = h$ and $he = fxee = fxe = h$. Thus one gets $hhe = he = h = hhh$. By unicity of the $h$-inverse of $h$, one has $h = e$ and thus $fe = e$ and symmetrically $ef = f$. Thus $e$ and $f$ are $\mathcal{R}$ equivalent.
Step 3: all the elements of $S$ are $\mathcal{R}$-equivalent. Let $x, y \in S$ and let $x'$ and $y'$ be inverses of $x$ and $y$, respectively. Then $x \mathrel{\mathcal{R}} xx'$, $y \mathrel{\mathcal{R}} yy'$, and since $xx'$ and $yy'$ are idempotent, $xx' \mathrel{\mathcal{R}} yy'$ by Step 2. Thus $x \mathrel{\mathcal{R}} y$.
Finally $S$ is a regular semigroup consisting of a single $\mathcal{R}$-class and thus is a right group.