What algebraic structure, if any, is determined by a uniquely left-solvable semigroup? By uniquely left-solvable semigroup I mean a set $\mathcal{S}$ with a mapping $f:\mathcal{S}\times\mathcal{S}\to\mathcal{S}$, represented by juxtaposition $f\left(a,b\right)\equiv{ab},$ such that $\forall_{a,b,c}\left(ab\right)c=a\left(bc\right),$ and $\forall_{a,b}\underline{\exists}_x xa=b.$  Here the notation $\underline{\exists}_x$ is to be read there exists exactly one $x$.  Right-solvable would mean $\forall_{a,b}\exists_x ax=b.$  Right-solvability is neither asserted nor precluded.
I believe this will prove to be an Abelian group, but I have not yet shown it to be so.  I am still working on it.  
This is completely for my own edification, and is not a homework problem, nor an exercise given in a book or other lesson.  I simply find the question interesting, and would like to share it.
 A: The condition does not define a group. Given any nonempty set $S$ with more than one element, define the product as $ab = a$ for all $a,b\in S$. This is a semigroup, and in this semigroup, the equation $xa=b$ has the unique solution $x=b$ for all $a$ and $b$; yet because $S$ has more than one element but every element is idempotent ($bb=b$ for all $b$), it cannot be a group.
(The comment that you hope it would define an abelian group is also incorrect, since any group satisfies the condition, abelian or not.)
As to whether this condition defines a specific, known class of semigroups, I don’t know (I’m not a semigroup theorist), but I can tell that it is a subclass of right-cancellable semigroups (semigroups in which $xa=ya$ implies $x=y$), but is not equal to the full class. Right cancellability follows by the uniqueness condition of the solution (in fact, saying “every equation has a unique solution” is equivalent to saying every equation has a solution and we have right-cancellability). However, there are right-cancellable semigroups that do not satisfy the condition, because not all such equations have solutions.
For a concrete example, let $X$ be an infinite set, and consider the subsemigroup of $T_X$ (the full transformation semigroup of all function $X\to X$ under composition) given by the surjective functions. This is a subsemigroup, since the composition of surjective functions is surjective. In addition, since surjective functions have right inverses/are right cancellable, this is a right-cancellable semigroup. However, not every equation $xa=b$ has a solution. For example, taking $b$ to be the identity, $xa=b$ has a solution if and only if $a$ is invertible (i.e., a bijection), so taking any non-injective surjection for $a$ gives an equation with no solution in this semigroup.
