The group $(\mathbb{Z},+)$ is a counterexample, of course, but that is probably not what the OP had in mind. In fact, the following theorem holds:
Theorem 1. Let $(G,\cdot)$ be a totally ordered Archimedean group. Then $(G,\cdot)$ is isomorphic to a subgroup of $(\mathbb{R},+)$ (which carries the order induced from $\mathbb{R}$). In particular, $(G,\cdot)$ is commutative.
This can, for instance, be found in [László Fuchs: Partially Ordered Algebraic Systems (1963), Theorem 1 in Section IV.1]; there, the theorem is attributed to Hölder.
Now, let $(H, +)$ be a subgroup of $(\mathbb{R},+)$. If the induced order on $H$ is dense (i.e. for all $a,c \in H$ which fulfil $a < c$ there exists $b \in H$ such that $a < b < c$), then it is not difficult to show that $H$ is also dense in $\mathbb{R}$ (note, however, that this observation relies on the fact that $(H,\cdot)$ is a subgroup of $(\mathbb{R},+)$; there are non-dense subsets of $\mathbb{R}$ which are densely ordered with respect to the induced order!).
If the order on $H$ is dense and complete, then we conclude that $H = \mathbb{R}$. This proves the following result:
Corollary 1. Let $(G,\dot)$ be a totally ordered Archimedean group whose order is dense and complete. Then $(G,\cdot)$ is isomorphic to $(\mathbb{R},+)$.
Remark: By isomorphic I mean that there exists a group isomorphism $\varphi$ such that both $\varphi$ and $\varphi^{-1}$ are increasing.
