Errata corrige. My "proof" that $(X,\tau)$ isn't semimetrizable when $X$ is countably infinite and $\tau$ is the cofinite topology on $X$ was flawed, and in fact, the contrary is true! This follows from a (straightforward) generalization of a theorem by W. A. Wilson dating back to the 1930s, which appears, e.g., as Theorem 6.3.50 in J. Goubault-Larrecq, Non-Hausdorff Topology and Domain Theory: Selected Topics in Point-Set Topology (New Math. Monographs 22, Cambridge Univ. Press, 2013), or can be recovered as an instance of a theorem by A. H. Frink on countably-based quasi-uniformities [...]

I'm considering monoids (to me, a monoid is a unital semigroup, or, if you prefer, a unital associative magma in the sense of Bourbaki). But I'm not sure to get the point of your question! What do you mean? Btw, my use of "in addition" in the present formulation of Q2 is misleading: every cancellative monoid is resilient.

Sorry for the late reply (I was on holiday). In fact, there're 1st-countable topologies which are not semimetrizable (to be honest, I had no doubt about their existence, but it's only two days ago that I found a counterexample): this is the case, e.g., with the cofinite topology on a countably infinite set. So yes, I restated the OP to take into account your comments.

@Chris: Thanks for your clarifications, which change many things and prove that also my 2nd reading of your answer was incorrect. To be honest, it wasn't so clear (to me?) which open sets you were actually adding to $\tau_X$: there's not a unique way to do it, and the "cheapest" way to do it is to add just $M$. Further, your answer still says: "Let $X$ be any topological space. [...]" So it's not really a question of being complicated or not. Btw, there's a mistake in my last comment: it should really read as "$d(0,x):=d(1,x):=0$ for all $x\in M$ and $d(x,0):=d(x,1):=\infty$ for $x\in X$".

Here's a "natural" follow up: mathoverflow.net/questions/163559.

If my 2nd reading of this answer (based on Eric's comments) is right, then Chris' construction doesn't however work, at least in general (and it's not clear to me if it works at all). Eric is in fact right: a semimetric $d_X$ on $X$ always extends to a semimetric $d$ on $M$ by $d(0,x):=d(x,0):=d(1,y):=d(y,1):=\infty$ for all $x,y\in M$ with $x\ne 0$ and $y\ne 1$; but the only neighborhood of both $0$ and $1$ in the canonical topology of $d$ is $M$, so $d$ induces $\tau_X\cup\{M\}$ if $d_X$ induces $\tau_X$ (the topology initially given on $X$). The rest follows from the previous comment.

@Eric Wofsey. I see! So I was misreading Chris' answer. He's just starting with a topology $\tau_X$ on $X$ and taking $\tau=\tau_X\cup\{M\}$ as a topology on $M$, right? That's fine! Still, how to prove that $\tau$ is not induced by a left/right $\mathbb M$-invariant semimetric? This boils down to $\tau$ not being semimetrizable at all (which is close to your remarks), since for a semimetric $d$ on $M$ we would have (for Chris' construction) that, for all $x,y,z\in M$, $d(xz,yz)=d(zx,zy)=d(0,0)\le d(x,y)$ if $z\ne 1$ or $d(xz,yz)=d(zx,zy)=d(x,y)$ if $z=1$, i.e. $d$ would be subinvariant.

(...) for which $\mathbb M$ is the unitization of a null sgrp, how do you think to prove that there doesn't exist any left/right $\mathbb M$-subinvariant semimetric $d$ on $M$ such that $\tau$ the canonical topology induced by $d$? That's the thing.

(...) $(\mathbb M,\tau)$ is not a topological monoid, unless $\{S\}\in\tau$, which is the case iff $S=\{0\}$. In fact, $M\setminus \{0\}$ is an open neighborhood of $1$ in $\tau$ (by construction), and given any open neighborhood $U$ of $(S,S)$ in the product topology induced on $M\times M$ by $\tau$ we should have that $xy\ne 0$ for all $x,y\in U$, but this is false unless we can take $U=\{(S,S)\}$, which in turn is possible iff $M\setminus \{0\}=\{S\}$, i.e. $S=\{0\}$. Do I miss something? And in any case, even assuming that $(\mathbb M,\tau)$ is a 1st-countable topological monoid (...)

[...] the discrete topology on $S$, as far as $S$ is countable, or more generally the topology $\{\emptyset, S\setminus \{0\}, S\}$); it is clear that $(\mathbb S,\tau_S)$ is a topological sgrp. Next, take $\mathbb M$ to be the unitization of $\mathbb S$, so that $\mathbb =S\cup\{S\}$ (I do everything in TG) and $\cdot$ is, by abuse of notation, the unique extension of the composition law of $\mathbb S$ to a binary operation on $M$ for which $xS=Sx=x$ for all $x$. Lastly, let $\tau=\tau_S\cup\{M,M\setminus\{0\}\}$. Then $\tau$ is a 1st-countable topology on $M$; however, (...)

Thank you, Eric, I will keep it in mind. It was yesterday that I discussed the problem with Jacek. We didn't get an answer, but he made a cute observation on the multiplicative structure of the real field, and this morning I ended up with the above.

I had thought of something in the same lines. Yet, I don't see how this answers my question. Essentially, you're taking as $\mathbb M$ the (forced) unitization of a null sgrp. Then, you're claiming that any topology on $M$ is compatible with the structure of $\mathbb M$. But I'm afraid this is false, even assuming that $\tau$ is 1st-countable. To see why, let $\mathbb S$ be a null sgrp with $\mathbb S=(S, \cdot)$, and let $\tau_S$ be any 1st-countable topology on $S$ such that $S \setminus \{0\} \in \tau_S$, where $0$ is the absorbing element of $\mathbb S$ (e.g., $\tau_S$ can be [...]