Semigroup with exactly one left(right) identity? Are there any examples of a semigroup (which is not a group) with exactly one left(right) identity (which is not the two-sided identity)? Are there any “real-world” examples of these (semigroups of some more or less well-known mathematical objects) or they could only be “manually constructed” from abstract symbols (a, b, c…) subject to operation given by a Caley table?
 A: Take the semigroup $S = \{a, b, 0\}$ with $a^2 = a$, $ab = b$ and every other product equal to $0$. Then $a$ is a left identity (since $ax = x$ for all $x \in S$) but it is not an identity (since $ba = 0$). You can define this semigroup in three other equivalent ways:
(1) As a transformation semigroup on $\{1, 2, 0\}$. Just take the semigroup generated by $a = [1, 0, 0]$ and $b = [2, 0, 0]$.
(2) As a semigroup of matrices. Just take $a = \pmatrix{1 & 0\\ 0 & 0}$, $b = \pmatrix{0 & 1\\ 0 & 0}$ and $0 = \pmatrix{0 & 0\\ 0 & 0}$.
(3) As the syntactic semigroup of the regular language $a^*b$ [or as the transformation semigroup of its minimal automaton, which is just (1)].
A: Consider for example the semigroup consisting of all constant functions on a set $X$ [acting on the right], together with one non-constant idempotent function $f$ (for example, let $f$ fix some point $x\in X$ and send every other point to some $y\neq x$).  Then $f$ is a unique left identity, and $f$ is not a right identity.
In general I think it's probably helpful to think about this question in terms of transformation semigroups.
EDIT:  Since this question has been sitting around with no accepted answer for a while, I'll state my last sentence a bit more strongly:  You can determine exactly which transformation semigroups have a single left (or right) identity, and since every semigroup is isomorphic to a transformation semigroup, doing this will give you all examples.
[Although I just noticed that the OP hasn't been on this site for about a month, so I guess the question might remain 'unanswered'.]
A: Take a finite semigroup $S$. Then $S$ has an idempotent element $e$ since $S$ is finite.
Let $T = \{se : s \in S\}$. Then $T$ is a subsemigroup of $S$. We have $e \in T$ because $e = ee$. And $e$ is a right identity of $T$ since $(se)e = s(ee) = se$ for all $s \in S$.
My problem with this example is that I don't think $e$ is the only right identity of $T$ for every such $T$ and this is probably still not real world enough.
I believe if we give more conditions when we construct $T$, we might be able to get a semigroup with only one right identity. But, it may not be a real example for OP.
