Does a non-abelian semigroup without identity exist?

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.

Take the set of all finite, non-empty sequences of elements in some non-empty set $S$ with operation given by concatenation (commonly called the free semigroup on $S$). That is, take $S^+=\bigcup_{n\geq 1}{S^n}$ where for $(s_1,\ldots, s_n),(t_1,\ldots, t_m)\in S^+$ we have their concatenation given by $$(s_1,\ldots, s_n)(t_1,\ldots, t_m)=(s_1,\ldots, s_n,t_1,\ldots, t_m).$$ This is non-commutative so long as $S$ contains at least two elements; if $x,y$ are such distinct elements, then $(x)(y)=(x,y)\neq (y,x)=(y)(x)$.

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.)