This is a follow up of Question 163246. For the reader's convenience, let me first copy&paste some basic definitions.

We let a semimetric on a set $X$ be a function $d: X \times X \to [0,\infty]$ such that $d(x,x) = 0$ and $d(x,z) \le d(x,y) + d(y,z)$ for all $x,y,z \in X$. Exactly as with "classical" metrics, $d$ induces a topology on $X$, which we call the canonical topology of $d$, having as a base the sets $\{y \in X: d(x,y) < r\}$ as $x$ ranges over $X$ and $r$ over $\mathbf R^+$.

Next, we take a topological monoid to be, as usual, a pair $(\mathbb M, \tau)$ consisting of a (multiplicatively written) monoid $\mathbb M = (M, \cdot)$ and a topology $\tau$ on $M$ such that $\cdot$ is continuous [...] in the expected sense.

Finally, if $\mathbb M = (M, \cdot)$ is a monoid and $d$ is a semimetric on $M$, we say that $d$ is: right $\mathbb M$-subinvariant if $d(xz,yz) \le d(x,y)$ for all $x,y,z \in M$; left $\mathbb M$-subinvariant if it is right $\mathbb M^{\rm op}$-subinvariant, where $\mathbb M^{\rm op}$ is the dual of $\mathbb M$; and $\mathbb M$-subinvariant if it is both right and left $\mathbb M$-subinvariant.

With that said, let $(\mathbb M, \tau)$ be a first-countable topological monoid, with $\mathbb M = (M, \cdot)$. It has been already observed in the other thread that $\tau$ does not need be the canonical topology of a left or right subinvariant semimetric. However, both of the answers there (my feeling is that Chris Schommer-Pries' construction can be made to work for a suitable topology, even if for the moment I don't know how) do "critically" rely on the fact that $\mathbb M$ has an absorbing element.

Based on the above, we then say that $\mathbb M$ is resilient if there doesn't exist any non-trivial submonoid $S$ of $\mathbb M$ [...] with an absorbing element (in particular, this means that the only idempotent element of $\mathbb M$ is the identity). Then here is my question:

Q. If $\mathbb M$ is resilient, is it possible for $\tau$ to be the canonical topology of a left (respectively, right) $\mathbb M$-subinvariant semimetric?

I don't think so (which means, in particular, that I don't have a counterexample), but it's well possible that I'm missing something trivial (it wouldn't be a novelty...). So thanks in advance for any comment, hint, pointer, or whatever.