Not really an example (see the edit below), but somehow related to Greg Kuperberg's one: Let $\mathbb A = (A, +, \cdot, \preceq)$ be a strictly totally orderable semiring (*) and for a fixed integer $n \ge 1$ let $\mathcal M_n(A)$ denote the set of all $n$-by-$n$ matrices with entries in $A$, endowed with the usual operations of addition, say $+$, and row-by-column multiplication, say $\cdot$, induced by $\mathbb A$. Then, $(\mathcal M_n(A), +, \cdot)$ is itself a semiring. So now, let ${\rm U}_n(\mathbb A^+)$ (respectively, ${\rm L}_n(\mathbb A^+)$) be the subsemigroup of $(\mathcal M_n(A), \cdot)$ consisting of all and the only upper (respectively, lower) triangular matrices whose entries on and above (respectively, below) the main diagonal are positive in $\mathbb A$. Then, both $({\rm U}_n(\mathbb A^+), \cdot)$ and $({\rm L}_n(\mathbb A^+), \cdot)$ are strictly totally orderable semigroups.
Edit. Sorry, I've just realized that the question was focused on groups, and not on semigroups/monoids! But then let me profit from my lack of attention to turn the above into a question: Do $({\rm U}_n(\mathbb A^+), \cdot)$ and $({\rm L}_n(\mathbb A^+), \cdot)$ embed into strictly totally ordered groups in the case when $(A, +, \cdot)$ is unital and commutative?
(*) Here, a semiring is just a (possibly non-unital or non-commutative) ring whose additive monoid is not necessarily a group. A strictly totally ordered semiring is then a $4$-uple $(A, +, \cdot, \preceq)$ such that $(A, +, \cdot)$ is a semiring and $\preceq$ is a total order on $A$ such that
- $(A, +, \preceq)$ is a strictly totally ordered magma (in fact, a monoid), namely $x + z \prec y + z$ and $z + x \prec z + y$ for all $x,y,z \in A$ with $x \prec y$.
- $x \cdot z \prec y \cdot z$ and $z \cdot x \prec z \cdot y$ for all $x,y,z \in A$ with $x \prec y$ and $0 \prec z$, where $0$ stands for the identity of the monoid $(A,+)$.