I apologize for answering my own question.
Let $\mathcal K = (\mathbb K, \tau)$ be a T1 topological unital ring, with $\mathbb K = (K, +, \cdot)$, and let $\mathbb K_{(\cdot)}$ be the multiplicative monoid of $\mathbb K$, so $(\mathbb K_{(\cdot)}, \tau)$ is a T1 topological monoid.
Assume that $\tau$ is induced by a semimetric $d$ on $K$. This implies that $d(x,y) \ne 0$ for all distinct $x,y \in K$ (as mentioned in the comments to the OP, the topology induced by a semimetric is T1 iff the "distance" of two distinct points is non-zero). In particular, we have $d(0,1) \ne 0$.
Now suppose that $d$ is right (respectively, left) subinvariant in $\mathbb K_{(\cdot)}$. Then, $0 < d(0,1) \le d(0, x^n)$ for every $x \in \mathbb K^\times$, where $\mathbb K^\times$ is, as usual, the set of the units of $\mathbb K$.
This is however impossible if there exists at least one element $x \in \mathbb K^\times$ such that $0$ is a limit point, relative to $\tau$, of the $K$-valued sequence $(x^n)_{n \in \mathbf N}$, which is for instance the case when $\mathbb K$ is the real field and $\tau$ is the usual topology on $\bf R$.
More in general, the above shows that if $\tau$ is the topology induced by a non-trivial absolute value $|\cdot|$ of $\mathbb K$ and if $|x| \ne 1$ for some $x \in \mathbb K^\times$, then $\tau$ can not be the canonical topology of a left (respectively, right) subinvariant semimetric on $K$, in spite of being first-countable (which serves as a partial answer to one of the questions raised by @Wlodzimierz Holsztynski in the comments to the OP, insofar as $d$ has no way of being topologically equivalent to the canonical metric induced on $K$ by $|\cdot|$).
I'd like to publicly thank Jacek Jendrej for a fruitful conversation which paved the way to all of this (as far as I know, he's not a MO user, and that's why I'm providing an external link).