In chapter $\rm{III}$ of the "Topologie générale" of Bourbaki -- which is the chapter they dedicate to a brief introduction to generalities concerning topological groups -- I have encountered an interesting statement, in the problem section (to be more precise, it is exercise 28. a) of section 2). The statement of the problem is as follows:
Consider an arbitrary topological group $(G, \cdot, \mathscr{T})$ and a subset $S \in \mathscr{T}$ such that $S \leqslant_{\mathrm{Sg}} G$ -- in other words an open subsemigroup of $G$ -- and furthermore such that $1_G \in \overline{S}$. It then holds that $S=\mathring{\overline{S}}$.
There is an indication given, namely to prove that any $x \in \overline{S} \setminus S$ is a boundary point for the adherence $\overline{S}$ (I should say this is what I reinterpreted from the original which reads any $x \in \overline{S} \setminus S$ is a boundary point for $S$, a statement trivially true by definition and which serves no purpose as a hint; typographical errors are rare in the edition I own, but they do sometimes show up...).
$S$ being open, the inclusion $S \subseteq \mathring{\overline{S}}$ is trivially valid, so it is the reverse one with which we need be concerned. What I achieved is to show that given a point $x \in \mathring{\overline{S}}$, for any neighbourhood $U \in \mathscr{V}_{\mathscr{T}}\left(1_G\right)$ of the unity the intersection $U \cap S \cap x^{-1}S \neq \varnothing$ is not empty, result for which I used indeed both key assumptions that $S$ is open and that $1_G$ is adherent to $S$. However, this does not seem enough to finalise the argument.
As the topology of any topological group is uniformisable, the particular description $\overline{X}=\displaystyle\bigcap_{V \in \mathscr{V}_{\mathscr{T}}\left(1_G\right)}XV$ is valid for any subset $X \subseteq G$, yet I don't see how this observation could be put to use in the given context. Yet another remark (which, sadly, again doesn't seem to be of great help) is that upon introducing $T\colon=\mathring{\overline{S}}$, we have that $T \leqslant_{\mathrm{Sg}} G$ is another open subsemigroup such that $S \subseteq T$ and furthermore such that $\overline{S}=\overline{T}$. Again, my perception is perhaps too muddled to clearly see how these facts combined with the hypothesis $T \setminus S \neq \varnothing$ could successfully lead to a Reductio ad Absurdum.
What especially puzzles me about this assertion is that it is allegedly valid in the highest generality, for even the most vague and pathological topological groups. I would be very grateful to anyone who has detailed knowledge and experience with such phenomena (including references in the literature) and who would be willing to shed some light on this mystery.
P.S. In a general topological space $(X, \mathscr{T})$ the following construction is available. The map given by: $$\begin{align} \mathscr{T} &\to \mathscr{T}\\ U &\mapsto \mathring{\overline{U}} \end{align}$$ is increasing, inflationary ($U \subseteq \mathring{\overline{U}}$) and idempotent, in other words it constitutes a closure operator on the topology $\mathscr{T}$ ordered by inclusion. The fixed elements of this closure operator we shall call regular open sets of $\mathscr{T}$. In the case of the above problem, the contention is precisely that the subsemigroup $S$ is regular in this topological sense. Would this perspective be of any use for the problem at hand?
It turns out that the argument is not so difficult at all. Set $T\colon=\mathring{\overline{S}} \in \mathscr{T}$ and consider arbitrary $x \in T$. It is essential to notice that since $1_G \in \overline{S}$ we also have $1_G=1_G^{-1} \in \left(\overline{S}\right)^{-1}=\overline{S^{-1}}$, so the unity is also adherent to the open subset $S^{-1} \in \mathscr{T}$.
Since $x \in T$ we equivalently have $1_G \in x^{-1}T \in \mathscr{T}$, since any translate of an open subset remains open. It follows that $x^{-1}T \in \mathscr{V}_{\mathscr{T}}\left(1_G\right)$ is a neighbourhood of the unity and by virtue of the essential observation above we have $x^{-1}T \cap S^{-1} \neq \varnothing$. Setting $U\colon=x^{-1}T \cap S^{-1}$, it is clear that $U \in \mathscr{T}$ is open, since it is expressed as the intersection of two open subsets. This entails that $\varnothing \neq xU=T \cap xS^{-1} \in \mathscr{T}$ is a nonempty open subset, so we can fix a certain $a \in xU$. As it is clear that $xU \subseteq T \subseteq \overline{S}$ and that $xU \in \mathscr{V}_{\mathscr{T}}(a)$ is a(n open) neighbourhood of $a \in \overline{S}$, it follows that $xU \cap S \neq \varnothing$ and subsequently -- since $xU \subseteq xS^{-1}$ -- we infer that $xS^{-1} \cap S \neq \varnothing$.
Explicitly, this means there exist $s, t \in S$ such that $xs^{-1}=t$ whence $x=ts \in SS \subseteq S$. By virtue of the arbitrariness of $x$ we can conclude that $T \subseteq S$ and finally that $S=\mathring{\overline{S}}$.