Are there any interesting and natural examples of semigroups that

Step 1
Consider the following definition of a group:
Definition A semigroup S is said to be a group if the following hold:
1. There is an ${\displaystyle e\in S}$ such that ${\displaystyle ea=a}$ for all ${\displaystyle a\in S}$
2. For each ${\displaystyle a\in S}$ there is an element ${\displaystyle a^{-1}\in S}$ with ${\displaystyle a^{-1}a=e}$
At one point in my life, it seemed natural to ask what happens if we replace axiom 2 with the very similar axiom
2'. For each ${\displaystyle a\in S}$ there is an element ${\displaystyle a^{-1}\in S}$ with ${\displaystyle a{a}^{-1}=e}$.
It is a fun exercise to work out some of the consequences that result from this. Here are a few facts about a semigroup S which satisfies 1 and 2':
If e is the unique element of S satisfying axiom 1, then S is a group
If S has an identity (in the usual sense) then S is a group
The principal left ideal ${\displaystyle Sa=\{sa\mid s\in S\}}$ is a group for all ${\displaystyle a\in S}$, and in fact all principal left ideals of S are isomorphic as groups.
It is not difficult to find examples of such semigroups that are not groups. For example, consider the following set of ${\displaystyle 2\times 2}$ matrices (with matrix multiplication as the operation):
$$\{(\begin{array}{cc}a& b\\ 0& 0\end{array})|a,b\in \mathbb{R},a\ne 0\}$$
Or, an example that appears as exercise 30 in section 4 of Fraleigh's abstract algebra text: the nonzero real numbers under the operation ${\displaystyle \times }$ defined by ${\displaystyle a\times b=\left|a\right|b}$.
Certainly it is debatable whether or not semigroups satisfying axioms 1 and 2′ are "interesting" or "natural". But I guess I think they are. And, I am not the only one (or the first one, by a long shot!) to think this. See Mann, On certain systems which are almost groups (MR).

Step 1
Let ${\displaystyle (M,\cdot ,e)}$ be a monoid and let ${\displaystyle M^{\circ }}$ be the set of finite words constructed from M. For two words let's define operation ${\displaystyle \times }$ as point-wise application of ${\displaystyle \cdot }$, truncating according to the shorter word:
${\displaystyle (u_1,\cdots ,u_m)\times (v_1,\cdots ,v_n)=(u_1\cdot v_1,\cdots ,u_{min(m,n)}\cdot v_{min(m,n)})}$
Then ${\displaystyle (M^{\circ },\times )}$ is a semigroup, but it's not a monoid. Any candidate y for the unit element would have only a finite length, so for any x that is longer we'd have ${\displaystyle y\times x\ne x}$. The unit element would have to be an infinite sequence ${\displaystyle (e,e,\cdots )}$, but ${\displaystyle M^{\circ }}$
This example is not arbitrary, it is closely related to zipping lists and convolution.
Step 2
Update: A very simple example of a semigroup that is not a monoid is ${\displaystyle (\mathbb{Z},min)}$. While min is clearly associative, there is no single element in ${\displaystyle \mathbb{Z}}$ that would serve as the identity. (It is actually a homomorphic image of the previous example, mapping words to their length.)

I don't know if this counts as interesting, but a simple example is what C programmers know as the comma operator: Ignore the first argument and return the second.
Or written as multiplication rule: $ab=b$ for all $a,b\in S.$
Or written as multiplication rule: $ab=b$ for all $a,b\in S.$. This is easily shown to be a semigroup, but as long as there are at least two elements in S, this is not a monoid, as with neutral element e you'd have for all $a\in S$ the identity $a=e$.