Is any finite semigroup of this type a left monoid? Let $(S, \cdot, e)$ be a semigroup $(S, \cdot)$ with binary operation $e$ in which the identities $e(x, y)\cdot x\approx x$ and $e(x, y)\approx e(y, x)$ hold.
In this question I asked if any such semigroup is necessarily a left monoid. Example given to me by J.-E. Pin shows that this isn't true. Clearly, $(\mathbb{Z}, \min, \max)$ is not a left monoid but satisfies those identities.
A left monoid is a semigroup with left identity.
Since I couldn't find a finite semigroup like this which wouldn't be a left monoid, and I've tried checking in GAP semigroups of order $\leq 4$, I suspect all finite semigroups of this form are left monoids for some combinatorial reasons.
Unfortunately, I'm not sure how to get all semigroups of orders, say, $\leq 7$, which would not be left monoids and be lwr semigroups, other than taking all the semigroups which aren't monogenic or monoids using GAP's Smallsemi package and checking if they are of this form by hand by creating a multiplication table. As you can imagine, this is very tedious.
Does there exist a finite semigroup of this form, not being a left monoid, and if yes, can you provide an example of least order?
 A: Any nonempty finite semigroup $S$ of this type has a left identity. First observe that for all $x, y \in S$,
$$
(1) \quad e(x,y)x = x \text{ and } e(x,y)y = e(y,x)y = y.
$$
Since $S$ is finite, it contains an idempotent $x_0$.
Let $S = \{x_0, x_1, \ldots, x_n\}$ and let $(a_i)_{0 \leqslant i \leqslant n}$ be the sequence of elements of $S$ defined by $a_0 = x_0$ and, for $1 \leqslant i \leqslant n$, $a_i = e(a_{i-1},x_i)$.
Claim: $a_n$ is a left identity of $S$.
First observe that, for $1 \leqslant i \leqslant n$,
\begin{align}
  &(2) \quad& a_ia_{i-1} &= e(a_{i-1},x_i)a_{i-1} = a_{i-1} \\
  &(3) \quad& a_ix_i &= e(a_{i-1},x_i)x_i = x_i 
\end{align}
Let us now prove by induction on $k = i+j$ that, for $0 \leqslant i \leqslant j$,
$$
(4) \quad a_jx_i = x_i. 
$$
If $k = 0$, then $i = j = 0$ and $a_0x_0 = x_0$ since $x_0$ is idempotent. Suppose that (4) holds for $i + j \leqslant k$ and suppose that $i + j = k+1$. If $i = j$, then (4) follows from (3). If now $i \leqslant j-1$, then $a_{j-1}x_i = x_i$ by the induction hypothesis. It follows by (2) that
$$
  a_jx_i = a_j(a_{j-1}x_i) = (a_ja_{j-1})x_i = a_{j-1}x_i = x_i.
$$
This proves the claim and concludes the proof.
