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 [Theorem 1 in Section IV.1](https://books.google.com/books?id=A_QxAwAAQBAJ&pg=PA45) of

<cite authors="Fuchs, L.">_Fuchs, L._, Partially ordered algebraic systems, Oxford-London-New York-Paris: Pergamon Press. IX, 229 p. (1963). [ZBL0137.02001](https://zbmath.org/?q=an:0137.02001).</cite>
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 $(H,+)$ 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 2.** Let $(G,\cdot)$ be a totally ordered Archimedean group whose order is dense and complete. Then $(G,\cdot)$ is isomorphic to $(\mathbb{R},+)$.

**Remark 3.** By *isomorphic* I mean that there exists a group isomorphism $\varphi$ such that both $\varphi$ and $\varphi^{-1}$ are increasing.


**-- Edit. --** In Corollary 2, the assumption that the group be Archimedean is redundant at least if the group is a priori known to be commutative. More precisely we have:

**Theorem 4.** Let $(G,+)$ be a commutative and totally ordered group. If the order on $G$ is complete, then $(G,+)$ is Archimedean.

*Proof.* Let $0 < b \in G$ and define $S := \{a \in G: \, a > 0 \text{ and } ma < b \text{ for all } m \in \mathbb{N}\}$. We have to show that $S$ is empty, so assume to the contrary that $a \in S$. Since $S$ is bounded above by $b$, $S$ has a supremum $s$ in $G$. Note that $s > 0$, so $2s > s$ and thus $2s \not\in S$, which in turn implies $s \not\in S$. Therefore, the set $S$ coincides with the interval $(0,s) := \{g \in G: 0 < g < s\}$.

In particular $s > a$, so $s-a \in (0,s) = S$. We now show that $s-a$ is actually *not* an element of $S$, which is a contradiction. As $s \not\in S$ we can find an integer $m \in \mathbb{N}$ such that $ms \ge b$. Hence,
\begin{align*}
  (m+1)(s-a) = ms + s - (m+1)a \ge ms \ge b;
\end{align*}
for the equality on the left we used that our group is commutative and for the inequality on the right we used that $(m+1)a \in S$ (since $a \in S$) and hence, $(m+1)a < s$. We have shown that $s-a \in S$, which is a contradiction.

**Remarks 5.** (a) I'm not sure whether Theorem 4 remains true if we do not assume $(G,+)$ to be commutative.

(b) Note that the existence of inverse elements is esssential not only for the proof, but also for the validity of Theorem 4. Indeed, the set $[0,\infty) \times [0,\infty)$, endowed with componentwise addition and the lexicographical order, is an example of a totally ordered semigroup (whose composition operation is strictly monotone in both components) which is order complete but not Archimedean.

**-- Edit 2 (a few hours later). --** Well, it seems that it is not too difficult to answer the question in Remark 5(a) above. So in order to complete the picture, let us add that Theorem 4 is actually true for non-commutative groups, too - and the proof is almost the same. Here are the details:

**Theorem 6.** Let $(G,\cdot)$ be a totally ordered group. If the order on $G$ is complete, then $(G,\cdot)$ is Archimedean.

*Proof.* Let $e \in G$ denote the neutral element of $G$. Let $e < b \in G$ and define $S := \{a \in G: \, a > e \text{ and } a^m < b \text{ for all } m \in \mathbb{N}\}$. We have to show that $S$ is empty, so assume to the contrary that $a \in S$. Since $S$ is bounded above by $b$, $S$ has a supremum $s$ in $G$. Note that $s > e$, so $s^2 > s$ and thus $s^2 \not\in S$, which in turn implies $s \not\in S$. Therefore, the set $S$ coincides with the interval $(e,s) := \{g \in G: e < g < s\}$. In particular $s > a$, so $sa^{-1} \in (e,s) = S$. Moreover, since $s \not\in S$, we can find an integer $m \in \mathbb{N}$ such that $s^m \ge b$. 

Up to now, the proof was exactly the same as for Theorem 4. Now was distinguish two cases.

*First case:* $sa^{-1} \ge a^{-1}s$. In this case we have
\begin{align*}
  (sa^{-1})^{m+1} \ge a^{-(m+1)} s s^m \ge s^m \ge b
\end{align*}
since $s > a^{m+1}$ (as $a^{m+1} \in S$). Hence, $sa^{-1} \not\in S$, which is a contradiction.

*Second case:* $sa^{-1} < a^{-1}s$. In this case we have
\begin{align*}
  (sa^{-1})^{m+1} \ge s^m s a^{-(m+1)} \ge s^m \ge b,
\end{align*}
again since $s > a^{m+1}$. Hence, we arrive at the same contradiction $sa^{-1} \not\in S$ as in the first case. This proves the theorem.

As a consequence, we get the following stronger version of Corollary 2:

**Corollary 7.** Let $(G,\cdot)$ be a totally ordered group whose order is dense and complete. Then $(G,\cdot)$ is isomorphic to $(\mathbb{R},+)$.