Added: The main results do not in fact depend on $R$ having no zero divisors, though I had to modify the arguments just a little in order to show this. I’ve made the necessary modifications below; significantly changed material is inset, like this note.
$R$ will be throughout a commutative ordered ring with identity, and $R^+\triangleq\{r\in R:r>0\}$. Define the $R$-norm $|\cdot|$ and $R$-metric $d$ as in the question. For $x,r\in R$ let $B(x,r)=\{y\in R:d(x,y)
Prop. 1: The $R$-metric topology on $R$ coincides with the order topology.
Proof: If $a, let $r=\min\{x-a,b-x\}>0$; clearly $x\in B(x,r)\subseteq(a,b)$. Conversely, $B(x,r)=(x-r,x+r)$. $\dashv$
Henceforth I assume that $R$ bears this topology.
Prop. 2: $R$ is discrete iff $R^+$ has a minimum element, and in that case $R$ is of course metrizable. $\dashv$
Assume henceforth that $R$ is not discrete.
Prop. 3: For each $x\in R^+$ there is a $y\in R^+$ such that $2y\le x$.
Proof: Let $x\in R^+$. Since $R$ is not discrete, there is a positive $u, and there is a positive $v. Let $y=\min\{u,v\}$; then $2y\le u+v=x$. $\dashv$ (This argument no longer depends on multiplication.)
Theorem 4: If there is a strictly decreasing sequence $\langle r_n:n\in\omega\rangle$ in $R^+$ that is co-initial in $R^+$, then $R$ is metrizable.
Proof: For $n\in\omega$ let $\mathscr{B}_n=\{B(x,r_n):x\in R\}$. Suppose that $V$ is an open nbhd of some $x\in R$. By Prop. 1 there is some $r\in R^+$ such that $B(x,r)\subseteq V$. By Prop. 3 there is an $n\in\omega$ such that $2r_n\le r$, whence $\operatorname{st}(x,\mathscr{B}_n)\subseteq V$. Thus, $\{\mathscr{B}_n:n\in\omega\}$ is a development for $R$. It’s well known that every linearly ordered topological space is collectionwise normal, so $R$ is metrizable by the Moore metrization theorem (also sometimes called the Bing metrization theorem). $\dashv$
Added: It takes a bit more work, but one can show that in this case $R$ has a metric $d$ such that the $R$-Cauchy sequences are precisely the ordinary Cauchy sequences in $\langle R,d\rangle$. (This is trivially true in the discrete case.)
Up to here, then, $R$-metrics don’t really give us anything new. However, there are non-metrizable ordered rings. The easiest example that comes to mind is a countable ultrapower of $\mathbb{Q}$.
Example 5: Let $X=\mathbb{Q}^\omega$, with addition and multiplication defined coordinatewise, and let $\mathscr{U}$ be a free ultrafilter on $\omega$. Let $I=\big\{x\in X:\{n\in\omega:x(n)=0\}\in\mathscr{U}\big\}$; $I$ is a maximal ideal in $X$, so $R=X/I$ is a field. For $[x],[y]\in R$ define $[x]\le[y]$ iff $\{n\in\omega:x(n)\le y(n)\}\in\mathscr{U}$; the fact that $\mathscr{U}$ is an ultrafilter ensures that this is a linear order on $R$, and it’s easily seen to be compatible with the ring structure on $R$.
Now suppose that $\{x_n:n\in\omega\}\subseteq R^+$.
Now suppose that $\{[x_n]:n\in\omega\}\subseteq R^+$; I’ll show that there is an $x\in R^+$ such that $x for every $n\in\omega$, thereby showing that $R$ is not first countable (and hence certainly not metrizable).
Let $U_0=\{k\in\omega:x_0(k)>0\}$. Given $U_n$ for some $n\in\omega$, let $U_{n+1}=U_n\cap\{k>n:x_{n+1}(k)>0\}.\,\tag{1}$ Since $[x_n]>[0]$ for each $n\in\omega$, each $U_n\in\mathscr{U}$. Clearly $U_0\supseteq U_1\supseteq\dots$, and $\bigcap\limits_{n\in\omega}U_n=\varnothing$. Now define $x\in X$ as follows:
$x(k)=\begin{cases} 0,&\text{if }k\in\omega\setminus U_0\\\\ \frac12\min\{x_m(k):m\le n\},&\text{if }k\in U_n\setminus U_{n+1}\;. \end{cases}$
It follows from $(1)$ that if $k\in U_n\setminus U_{n+1}$, $x_m(k)>0\,$ for every $m\le n$ and hence that $x(k)>0$. Thus, $x(k)>0$ for every $k\in U_0\in\mathscr{U}$, and hence $[x]>[0]$. On the other hand, for each $n\in\omega$ we have $x(k) for every $k\in U_n\in\mathscr{U}$, so $[x]<[x_n]$. $\dashv$
Definition 6: A sequence $\langle x_n:n\in\omega\rangle$ is $R$-Cauchy iff for each $r\in R^+$ there is an $n_r\in\omega$ such that $d(x_n,x_m) whenever $n,m\ge n_r$.
Obviously if $R$ is discrete, $R$-Cauchy sequences are eventually constant, while in the non-discrete metrizable case they behave like ordinary Cauchy sequences in metric spaces.
Prop. 7: If $R$ is not first countable, a sequence in $R$ is $R$-Cauchy iff it is eventually constant.
Proof: Clearly $R$ has a nested base at $0$, and this base cannot be countable: if it were, $R$ would be first countable. Let $\langle x_n:n\in\omega\rangle$ be an $R$-Cauchy sequence in $R$, and let $D=\{d(x_n,x_m):n,m\in\omega\}\cap R^+,$ the set of non-zero distances between terms of the sequence. $D$ is countable, so there is an $r\in R^+$ such that $r for every $d\in D$. Fix $n_r\in\omega$ such that $d(x_n,x_m) whenever $n,m\ge n_r$; clearly $d(x_n,x_m)=0$ and $x_n=x_m$ whenever $n,m\ge n_r$, i.e., the sequence is eventually constant. $\dashv$
Cor. 8: If $R$ is not discrete, $R$ is metrizable iff it has an $R$-Cauchy sequence that is not eventually constant. $\dashv$