Let $\mathcal{M}=(M,0,+,<)$ be a linearly ordered divisible abelian group. Let's define an $\mathcal{M}$-metric on $M$ to be a map $d:M\times M \rightarrow M$ such that
(1) $\forall x,y\in M,\, d(x,y)=0 \Leftrightarrow x=y$,
(2) $\forall x,y\in M,\, d(x,y)=d(y,x)$,
(3) $\forall x, y, z\in M, \, d(x,y) \leq d(x,z) + d(z,y)$.
Note that one such map defines a topology on $M$ in the usual way that metrics do.
Let $\tau$ be the right half-open interval topology in $M$. Namely the topology with basis $\{ [x,x+\varepsilon) : \, x, \varepsilon\in M, \varepsilon>0\}$.
In there some $\mathcal{M}$-metric on M that generates $\tau$?
If $\mathcal{M}=(\mathbb{R},0,+,<)$ then the answer is no since the topological space described would be separable and not second countable, hence not metrizable.
Any ideas/intuitions regarding the question, even if they are not full answers, are apreciated.
To manage expectations, let me start with a simple observation. $M$ does not admit a translation-invariant $\mathcal{M}$-metric. If $\rho$ is a translation-invariant $\mathcal{M}$-metric, then $\rho(x, 0) = \rho(-x, 0)$ for all $x$, which means that any open ball about $0$ is symmetric. Hence the open set $[0, \rightarrow)$ cannot contain such a ball. This suggests that even if a metric exists, it may not be very useful.
The same reasoning applied to $\mathbb{R}$ can be applied to many other cases. If $M$ admits an $\mathcal{M}$-metric, then $d(M) = |M|$, i.e. $M$ does not have a dense subset of lower cardinality. Since the balls with centre and radius in a dense subset form a base for the topology, we have $w(M) \le d(M)d(M) = d(M)$. On the other hand we have $w(M) \ge |M|$ for the same reason as in $\mathbb{R}$. (Note that here we do not have to distinguish between order-dense, dense in the order topology and dense in the half-open interval topology, as they are equivalent for a dense linear order.)
This necessary condition can be sharpened by an observation that is useful for non-Archimedean groups. If $\mathcal{N}$ is a nontrivial convex subgroup of $\mathcal{M}$, the following are equivalent:
That 4. implies 1. is trivial and 1. implies 2. because the restriction of a metric to a subspace induces the subspace topology. The implication $2. \implies 3.$ can be obtained by taking a positive $b \in\mathcal{N}$ and defining $\sigma(x, y) = \min\{\rho(x,y), b\}$ as usual. Since $N$ is open in $M$, the quotient $M/N$ is discrete and $M$ is homeomorphic to $(M/N) \times N$. Hence, given an $\mathcal{N}$-metric on $N$, we can make $M$ locally isometric to $N$ to arrive at 4.
Giving a useful necessary and sufficient condition is probably very hard, but there is a sufficient condition that is not too difficult. If $\mathcal{M}$ has a nontrivial countable convex subgroup, $M$ admits an $\mathcal{M}$-metric. By the previous result, we need only consider the case $|M| = \aleph_0$. Let $B$ be the set of real numbers between 0 and 1 whose binary expansions have finitely many ones. This is also a countable dense linearly ordered set with no minimum or maximum, so it is order-isomorphic to $M$. It will suffice then to find an $\mathcal{M}$-metric that induces the half-open interval topology on $B$.
Since $M$ is countable, there is a strictly decreasing sequence $\{a_n\}$ of positive elements with $\inf_n a_n = 0$. Define $\rho(x, y) = 0$ when $x=y$ and otherwise $\rho(x, y) = a_m$, where $m$ is the largest integer such that the binary expansions of $x$ and $y$ are equal to $m$ places. Symmetry and positivity of $\rho$ are obvious and it is easily seen to satisfy the ultrametric triangle inequality $\rho(x, z) \le \max\{\rho(x, y),\rho(y,z)\}$.
To see that the topology induced by $\rho$ is finer that the half-open interval topology, note that in the latter a point $x$ has a base of neighbourhoods of the form $[x, y)$ where $y>x$. There is an $n$ such that the binary expansions of $x$ and $y$ contain no ones after the $n$th place. For such $n$, the $\rho$-ball of radius $a_{n+1}$ about $x$ is contained in $[x, y)$. Similarly, the topology induced by $\rho$ is coarser than the half-open interval topology, since every $\rho$-ball about $x$ contains a half-open interval starting at $x$. Thus $\rho$ induces the half-open interval topology on $B$.
Note that under the same hypothesis, $M$ also admits a real-valued metric, by Urysohn's metrization theorem.
Side note: For topological purposes, it usually makes no difference whether a ordered abelian group is divisible or just densely ordered. On one hand, in a densely ordered group there is for every $b > 0$ and every positive integer $n$ an $a > 0$ such that $na <= b$. On the other hand, in a divisible group there may be no $a > 0$ such that $\lim_{n\to\infty} a/n = 0$.