The linearly ordered group $(\mathbb{Z},+,\le)$ is a counterexample, but that is probably not what the OP had in mind. To give a detailed description of the situation, let us use the following notation:

By a *linearly bi-ordered group* we mean a tuple $(G,\cdot,\le)$ where $(G,\cdot)$ is a group and $\le$ is a linear order on $G$ such that $ac \le bc$ and $ca \le cb$ whenever $a,b,c \in G$ such that $a \le b$. We use the notion *linearly ordered group* as shorthand or *linearly bi-ordered group*.

An *isomorphism* between two linearly ordered groups $(G,\cdot,\le)$ and $(H,\cdot,\le)$ is a group isomorphism $\varphi: (G,\cdot) \to (H,\cdot)$ such that both $\varphi$ and $\varphi^{-1}$ are increasing.

A linearly ordered group $(G,\cdot,\le)$ (whose neutral element we denote by $e$) is called *Archimedean* if, for all $a,b > e$ there exists an integer $n \in \mathbb{N}$ such that $a^n \ge b$.

We call a linear order an a set $S$ *complete* if every non-empty subset of $S$ that is bounded above has a supremum in $S$ (equivalently, every non-empty subset of $S$ that is bounded below has an infimum in $S$). Note that this property is sometimes called *conditionally complete* (instead of *complete*) in the literature.

**Theorem 1.** Let $(G,\cdot,\le)$ be an Archimedean linearly ordered group. Then $(G,\cdot)$ is isomorphic to an ordered subgroup of $(\mathbb{R},+,\le)$ (i.e. a subgroup of $(\mathbb{R},+)$ which carries the order inherited from $\mathbb{R}$). In particular, $(G,\cdot)$ is commutative.

This result can, for instance, be found in Theorem 1 in Section IV.1 of

*Fuchs, L.*, Partially ordered algebraic systems, Oxford-London-New York-Paris: Pergamon Press. IX, 229 p. (1963). ZBL0137.02001.
There, the theorem is attributed to Hölder.

As kindly pointed out by user Alec Rhea in the comments, there is a related result by Hahn which gives a description of all *commutative* linearly ordered groups.

Next we note that linearly ordered groups whose order is complete are automatically Archimedean:

**Theorem 2.** Let $(G,\cdot,\le)$ be a linearly ordered group and assume that the order $\le$ on $G$ is complete. Then $(G,\cdot,\le)$ is Archimedean.

*Proof.* Let $e$ denote the neutral element of $(G,\cdot)$, let $a,b > e$ and assume for a contradiction that $a^n < b$ for all $n \in \mathbb{N}$. Then the set $S := \{a^n: \, n \in \mathbb{N}\}$ has a supremum $s$ in $G$. We have $a^{-1}s < s$, so $a^{-1}s < a^n$ for some $n \in \mathbb{N}$. Consequently, $s < a^{n+1} \le s$, which is a contradiction.

In an earlier version of this post, a more complicated proof of Theorem 2 was given. The above version of the proof was kindly pointed out by user Emil Jeřábek in the comments.

**Remark 3.** Note that the existence of inverse elements is esssential not only for the proof, but also for the validity of Theorem 2. Indeed, the set $[0,\infty) \times [0,\infty)$, endowed with componentwise addition and the lexicographical order, is an example of a linearly ordered semigroup (whose composition operation is strictly monotone in both components) which is order complete but not Archimedean.

By combining Theorems 1 and 2 we arrive at the following corollary which, I think, answers the question of the OP:

**Corollary 4.** Let $(G,\cdot,\le)$ be a linearly ordered group. If the order $\le$ on $G$ is complete, then $(G,\cdot,\le)$ is isomorphic to one of the three linearly ordered groups $(\{0\},+,\le)$, $(\mathbb{Z},+,\le)$ and $(\mathbb{R},+,\le)$.

*Proof.* According to Theorem 2 $(G,\cdot,\le)$ is Archimedean, so it is isomorphic to an ordered subgroup $(H,+,\le)$ of $(\mathbb{R},+,\le)$ due to Theorem 1. If $H$ has only one element, then obviously $H = \{0\}$, so assume that $H$ has at least two elements. Now we distinguish two cases:

*First case: $h_0 := \inf \{h \in H: \, h > 0\} > 0$.* Then it is easy to see that $H = h_0 \mathbb{Z}$, so $(H,+,\le)$ is isomorphic to $(\mathbb{Z},+,\le)$.

*Second case: $\inf \{h \in H: \, h > 0\} = 0$.* Then one readily checks that the set $H$ is dense in $\mathbb{R}$, and the completeness of the order on $H$
implies that $H = \mathbb{R}$. This proves the corollary.

**-- Note on edits made 2018-06-10. --** I rewrote the answer and consolidated the various edits from the previous versions in order to make this post better readable for future visitors. I also incorporated various suggestions by users Alec Rhea, Emil Jeřábek and YCor, so let me thank them for their comments!