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I'd like to prove that we can get rid of Conditions 2 and 4 from your list of wishes, for then, in order to conclude, it will be enough to consider your preferred finite commutative band and extend it, if necessary, according to the steps below.

For, suppose that we have somehow built a semigroup $\mathbb A = (A, \cdot)$, either finite or infinite, for which Conditions 1 and 3 hold true. First, we unitize $\mathbb A$ by forcing an identity even if $\mathbb A$ is already unital. It is seen that the outcome is still a semigroup, let me denote it by $\mathbb A^{(1)}$ for which Conditions 1 and 3 continue to be true (unless I'm missing something). More interestingly, we have a gain in the process, because both of the regular representations of $\mathbb A^{(1)}$ are now faithful (in the sense of your definition). This shows that we can assume from the outset that $\mathbb A$ is a monoid fulfilling Conditions 1, 3 and 4 (as for the latter, even in a stronger form).

At this point, pick an element $\infty \notin A$ and extend $A$ to the set $S := A \cup \{\infty\}$ and $\cdot$ to a binary operation on $S$, here still denoted by $\cdot$, by taking $\infty \cdot \infty := e$ and $\infty \cdot x := x \cdot \infty := x$ for $x \in S$, where $e$ is the identity of $\mathbb A$. Then, $(S, \cdot)$ is still a semigroup for which Conditions 1, 3 and 4 are verified (again, unless I'm missing something), but in addition $S \cdot S = A \subsetneq S$.

Added later. As for a concrete example (based on the above), we can just consider a set $S$ with three elements $1$, $e$ and $\infty$ and let $\cdot$ be the binary operation on $S$ given by the following (Cayley) table:

$$\begin{array}{c|ccc} \cdot & 1 & e & \infty \\ \hline 1 & 1 & e & 1 \\ e & e & e & e \\ \infty & 1 & e & e \end{array}.$$ The recipe is: We start with the trivial group with carrier $\{1\}$, then we unitize it by forcing a new identity $e \ne 1$, and lastly we adjoin a further element $\infty \notin \{1, e\}$ and extend the multiplication table by taking $\infty \cdot \infty := e$ and $\infty \cdot x = x \cdot \infty = x$ for $x = 1$ or $e$.

Not an answer, but a question based on a previous attempt of finding a solution, for which Benjamin Steinberg pointed out a foolish mistake in the comments below. Is there an "enlarge-and-shrink recipe" to extend a semigroup $\mathbb A = (A, \cdot)$ for which Conditions 1, 3 and 4 hold true to a larger semigroup $(S, \cdot)$ for which Conditions 1, 3 and 4 continue to be true, but in addition $S^2 \ne S$? If the answer is yes, then the problem is solved in the positive: Start with your preferred self-dual band $\mathbb A$, unitize it by adjoining an identity only if $\mathbb A$ is not already unital (in such a way that the unitization is still a self-dual band, but we have a gain in the process, since now the outcome is a semigroup whose regular representations are both faithful, regardless as to whether or not this was already the case with $\mathbb A$), and finally use the enlarge-and-shrink recipe to conclude.

I'd like to prove that we can get rid of Conditions 2 and 4 from your list of wishes, for then, in order to conclude, it will be enough to consider your preferred commutative band and extend it, if necessary, according to the steps below.

First, suppose that we have somehow built a semigroup $\mathbb A = (A, \cdot)$, either finite or infinite, for which Conditions 1 and 3 hold true. First, we unitize $\mathbb A$ by forcing an identity even if $\mathbb A$ is already unital. It is seen that the outcome is still a semigroup, let me denote it by $\mathbb A^{(1)}$ for which Conditions 1 and 3 continue to be true (unless I'm missing something). More interestingly, we have a gain in the process, because both of the regular representations of $\mathbb A^{(1)}$ are now faithful (in the sense of your definition). This shows that we can assume from the outset that $\mathbb A$ is a monoid fulfilling Conditions 1, 3 and 4 (as for the latter, even in a stronger form).

At this point, pick an element $\infty \notin A$ and extend $A$ to the set $S := A \cup \{\infty\}$ and $\cdot$ to a binary operation on $S$, here still denoted by $\cdot$, by taking $\infty \cdot \infty := e$ and $\infty \cdot x := x \cdot \infty := x$ for $x \in S$, where $e$ is the identity of $\mathbb A$. Then, $(S, \cdot)$ is still a semigroup for which Conditions 1, 3 and 4 are verified (again, unless I'm missing something), but in addition $S \cdot S = A \subsetneq S$.

Edits. Sorry, I was writing while Boris posted his answer.

I'd like to prove that we can get rid of Conditions 2 and 4 from your list of wishes, for then, in order to conclude, it will be enough to consider your preferred finite commutative band and extend it, if necessary, according to the steps below.

For, suppose that we have somehow built a semigroup $\mathbb A = (A, \cdot)$, either finite or infinite, for which Conditions 1 and 3 hold true. First, we unitize $\mathbb A$ by forcing an identity even if $\mathbb A$ is already unital. It is seen that the outcome is still a semigroup, let me denote it by $\mathbb A^{(1)}$ for which Conditions 1 and 3 continue to be true (unless I'm missing something). More interestingly, we have a gain in the process, because both of the regular representations of $\mathbb A^{(1)}$ are now faithful (in the sense of your definition). This shows that we can assume from the outset that $\mathbb A$ is a monoid fulfilling Conditions 1, 3 and 4 (as for the latter, even in a stronger form).

At this point, pick an element $\infty \notin A$ and extend $A$ to the set $S := A \cup \{\infty\}$ and $\cdot$ to a binary operation on $S$, here still denoted by $\cdot$, by taking $\infty \cdot \infty := e$ and $\infty \cdot x := x \cdot \infty := x$ for $x \in S$, where $e$ is the identity of $\mathbb A$. Then, $(S, \cdot)$ is still a semigroup for which Conditions 1, 3 and 4 are verified (again, unless I'm missing something), but in addition $S \cdot S = A \subsetneq S$.

Added later. As for a concrete example (based on the above), we can just consider a set $S$ with three elements $1$, $e$ and $\infty$ and let $\cdot$ be the binary operation on $S$ given by the following (Cayley) table:

$$\begin{array}{c|ccc} \cdot & 1 & e & \infty \\ \hline 1 & 1 & e & 1 \\ e & e & e & e \\ \infty & 1 & e & e \end{array}.$$ The recipe is: We start with the trivial group with carrier $\{1\}$, then we unitize it by forcing a new identity $e \ne 1$, and lastly we adjoin a further element $\infty \notin \{1, e\}$ and extend the multiplication table by taking $\infty \cdot \infty := e$ and $\infty \cdot x = x \cdot \infty = x$ for $x = 1$ or $e$.

This is not a complete answer, but it is too long for a comment. I'd like to prove that we can get rid of Conditions 2 and 4 from your list of wishes.

For, suppose that we have somehow built a semigroup $\mathbb A = (A, \cdot)$, either finite or infinite, for which Conditions 1 and 3 hold true. First, we unitize $\mathbb A$ by forcing an identity even if $\mathbb A$ is already unital. It is seen (unless I'm missing something) that the outcome is still a semigroup, let me denote it by $\mathbb A^{(1)}$, for which Conditions 1 and 3 continue to be true. More interestingly, we gain something in the process, because both of the regular representations of $\mathbb A^{(1)}$ are now faithful (in the sense of your definition). This shows that we can assume from the outset that $\mathbb A$ is a monoid fulfilling Conditions 1, 3 and 4 (as for the latter, even in a stronger form).

At this point, pick an element $\infty \notin A$ and extend $A$ to the set $S := A \cup \{\infty\}$ and $\cdot$ to a binary operation on $S$, here still denoted by $\cdot$, by taking $\infty \cdot \infty := e$ and $\infty \cdot x := x \cdot \infty := x$ for $x \in S$, where $e$ is the identity of $\mathbb A$. Then, $(S, \cdot)$ is still a semigroup (again, unless I'm missing something), for which Conditions 1, 3 and 4 are verified, but in addition $S \cdot S = A \subsetneq S$.

Edit. Sorry, I was writing while Boris posted his answer.

I'd like to prove that we can get rid of Conditions 2 and 4 from your list of wishes, for then, in order to conclude, it will be enough to consider your preferred commutative band and extend it, if necessary, according to the steps below.

First, suppose that we have somehow built a semigroup $\mathbb A = (A, \cdot)$, either finite or infinite, for which Conditions 1 and 3 hold true. First, we unitize $\mathbb A$ by forcing an identity even if $\mathbb A$ is already unital. It is seen that the outcome is still a semigroup, let me denote it by $\mathbb A^{(1)}$ for which Conditions 1 and 3 continue to be true (unless I'm missing something). More interestingly, we have a gain in the process, because both of the regular representations of $\mathbb A^{(1)}$ are now faithful (in the sense of your definition). This shows that we can assume from the outset that $\mathbb A$ is a monoid fulfilling Conditions 1, 3 and 4 (as for the latter, even in a stronger form).

At this point, pick an element $\infty \notin A$ and extend $A$ to the set $S := A \cup \{\infty\}$ and $\cdot$ to a binary operation on $S$, here still denoted by $\cdot$, by taking $\infty \cdot \infty := e$ and $\infty \cdot x := x \cdot \infty := x$ for $x \in S$, where $e$ is the identity of $\mathbb A$. Then, $(S, \cdot)$ is still a semigroup for which Conditions 1, 3 and 4 are verified (again, unless I'm missing something), but in addition $S \cdot S = A \subsetneq S$.

Edits. Sorry, I was writing while Boris posted his answer.