Does a non-abelian semigroup without identity exist?
Solution 1:
If $S$ is a set with at least two elements, and multiplication is defined by $x\cdot y=y$, then $(S,\cdot)$ is a noncommutative semigroup with no right identity. (Of course every element is a left identity.)
Solution 2:
Let $A^{\ast}$ be the free monoid on a nonempty set $A$; this is just the set of all words using $A$ as an alphabet. Let $e$ be the identity. Since no element of $A^{\ast}$ has an inverse other than $e$, $A^{\ast}-\{e\}$ is a semigroup without identity, and it is nonabelian provided $A$ has more than one element.