Does such a semigroup exist? A homework question asks if there exists a semigroup that satisfies:
$\forall x\forall y \exists z (x\circ z=y)$, $\forall x\forall y \exists z' (z'\circ x=y)$, $\exists x \exists y\exists z(x\circ z=y \wedge z\circ x \neq y)$
All element has a way to any other element via left or right multiplication, but the left "intermediate" element from x to y is different from the right "intermediate", for some x and y.

Your homework question describes a kind of object called a "non-Abelian group". The smallest example is the set of all permutations of a $3$-element set $X$ (that is, all bijections $f:X\to X$) with the operation of composition. An example that may be familiar to you is the set of all invertible $2\times2$ matrices with the operation of matrix multiplication.
