A finitely based monoid whose semigroup reduct is not finitely based, and vice versa

Question

Does there exist a monoid $(M,*,1)$ which is finitely based, but whose $\{*\}$ reduct is not finitely based? Also, does there exist a monoid $(M,*,1)$ which is not finitely based, but whose $\{*\}$ reduct is finitely based?

Answer 1

Case 0. (Relevant, but not the question asked!) It is known that there is a finitely based semigroup $S$ such that the monoid $S^1$ is not finitely based (Peter Perkins, 1969). It is also known that there is a nonfinitely based semigroup $T$ such that the monoid $T^1$ is finitely based (Edmund Lee, 2013).

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But you are not asking about formally adjoining a unit to a semigroup. Let's turn to the question asked.

Case 1. $(M,*,1)$ is a monoid satisfying some identity of the form $x^n=1$, $n>0$.

In this case, if $\Sigma$ is a finite semigroup basis for $(M,*)$, then $\Sigma\cup \{x^n=1\}$ is a finite monoid basis for $(M,*,1)$. Conversely, if $\Gamma$ is a finite monoid basis for $(M,*,1)$, then a finite semigroup basis for $(M,*)$ is $\{x^ny=y, yx^n=y\}\cup \Gamma'$ where $\Gamma'$ is obtained from $\Gamma$ by replacing all instances of $1$ with $x^n$.

What seems to be the hard case is

Case 2. $(M,*,1)$ satisfies no identity of the form $x^n=1$.

In this case, we still have the following: if $\Sigma$ is a finite semigroup basis for $(M,*)$, then a finite monoid basis for $(M,*,1)$ is $\Sigma\cup \{x*1=x, 1*x=x\}$.

I don't see how to argue "conversely" in Case 2. That is, I don't see how to rule out the possibility that $(M,*)$ is nonfinitely based and $(M,*,1)$ is finitely based when $(M,*,1)$ satisfies no identity of the form $x^n=1$.