If the underlying order of a nontrivial ordered group is discrete (i.e. every element has a successor and a predecessor), then either the group is isomorphic to $\mathbb{Z}$ or it is non-archimedean. Indeed, if $1$ (the successor of $0$) does not generate the whole group, then there is some element which is greater than $n\cdot 1$ for all $n\in \mathbb{N}$. The same statement holds for ordered rings, since the only ordered ring structure on $\mathbb{Z}$ compatible with the ordered group structure is the standard one.
So for a theory of ordered groups or rings all of whose models are non-archimedean, we can take $\mathrm{Th}(M)$, where $M$ is any discretely ordered group or ring which is not elementarily equivalent to $\mathbb{Z}$. At this point, you might worry that the theory of discretely ordered groups or discretely ordered rings is complete, so every model is elementarily equivalent to $\mathbb{Z}$. Fortunately, that's not the case.
For example, consider $T = \text{Th}(\mathbb{Z}\times\mathbb{Z},0,+,-, <)$, where $\mathbb{Z}\times \mathbb{Z}$ is ordered lexicographically. The sentence asserting that for all $x$, either $x$ or its successor is divisible by $2$ is true in $\mathbb{Z}$ but false in $\mathbb{Z}\times\mathbb{Z}$ (neither $(1,0)$ or $(1,1)$ are divisible by $2$).