Edit: As Emil Jeřábek has pointed out in the comments, there is a stronger version of the proposition available that gives a more direct proof.

**Theorem**. If $(S, +, <)$ is an ordered abelian group whose order type is dense and complete, then $S$ is isomorphic to $(\mathbb{R}, +, <)$.

It follows immediately that no ordered group can be Suslin. Emil credits this to Hölder, and gives a proof of the theorem in his answer here: mathoverflow.net/a/179520/12705

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I worked through this problem with Geoff Galgon. For Suslin lines at least the answer is negative: no such line can be the order type of an ordered abelian group. A key point is the following:

**Proposition**. If $(S, +, <)$ is an ordered abelian group whose order type is dense and complete, then there is an embedding of $(\mathbb{R}, +, <)$ into $S$.

From this it follows that no such $S$ can be Suslin. For now we'll delay the proof of the proposition and give the remainder of the argument.

Suppose toward a contradiction that $(S, +, <)$ is a linearly ordered abelian group whose order type is Suslin. For every $c \in S$, the map $x \mapsto x+c$ is an order-automorphism of $S$. It follows that no interval $(a, b)$ of $S$ can be separable (where we allow $a=-\infty$ or $b=\infty$, though in those cases we need to modify the following slightly). If there were such an interval, then by translating to the left and right in increments of $b-a$ and using the fact that $S$ is ccc, we would be able to cover $S$ (minus some endpoints) with countably many copies of $(a, b)$ (though we may need more than $\omega$-many in each direction: if, say, $\omega$ copies of $(a,b)$ does not cover $S$ to the right, translate $(a,b)$ so that $a$ sits at the limit point of this sequence of intervals, which exists since $S$ is complete, and continue covering incrementally). This is impossible, as then $S$ would be separable.

Now, by our proposition there exists a subgroup of $S$ that is isomorphic to $\mathbb{R}$. This subgroup is cofinal and coinitial in some open interval around $0$. Since this interval is not separable, our copy of $\mathbb{R}$ cannot cover it completely. That is, there must be some cut in $\mathbb{R}$ that contains a point in $S \setminus \mathbb{R}$. This cut either has the form $(-\infty, r] \cup (r, \infty)$ or $(-\infty, r) \cup [r, \infty)$. We may assume it's of the former kind, and for simplicity let's suppose $r=0$. So there is at least one point in $S \setminus \mathbb{R}$ that sits above $0$ but below every real number greater than $0$. In fact, since the order type of $S$ is dense, there must be an entire interval $I_0 \subset S \setminus \mathbb{R}$ sitting in that region. We take $I_0$ to denote the maximal such interval, that is, all points in $S \setminus \mathbb{R}$ between $0$ and the positive reals. This interval has the form $(0, s]$, where $s$ is the infimum in $S$ of the positive reals.

Now, for any real $r \in \mathbb{R}$, consider the translation of $S$ by $r$. This is an order-automorphism of $S$ that restricts to an order-automorphism of $\mathbb{R}$, and it sends $0$ to $r$. Necessarily then, it sends the interval $I_0$ onto some interval $I_r$ above $r$ but below ever real greater than $r$ (indeed it sends it onto the maximal such interval). For $r < r'$, the intervals $I_r$ and $I_{r'}$ are disjoint, since if they were not we would have $r' \in I_r$, impossible as $I_r$ contains no reals.

But now we have a contradiction, since we have found an uncountable collection of disjoint intervals in $S$, namely $\{I_r: \, r \in \mathbb{R} \}$.

*Proof of the proposition*: It seems likely that similar arguments to the following have been done before. If anyone knows of standard results to cite that would streamline things, I would be glad to hear them.

Let $(S, +, <)$ be an ordered abelian group whose order type is dense and complete. Completeness is a strong hypothesis: we'll show that such an $S$ is actually a vector space over $\mathbb{Q}$. This goes by first showing, for every $s \in S$ and $n \in \mathbb{N}$, there is an element $y$ (necessarily unique), which we denote as $\frac{1}{n}s$, such that $ny=s$. We may then define $\frac{m}{n} s = m (\frac{1}{n}s)$, and check that these fractional coefficients satisfy the vector space axioms for scalars.

Once we've done this, we can show that $S$ contains the desired copy of the ordered group $\mathbb{R}$. First, fix any element greater than $0$ in $S$ and label it $1$. Let $Q$ denote the one-dimensional subspace of $S$ generated by $1$. Then $Q$ is isomorphic (as a linear order, group, subspace) to $\mathbb{Q}$. Since $S$ is complete, every unfilled cut in $Q$ contains at least one point, and possibly many. As such there may be many ways to take a Dedekind completion of $Q$ within $S$ to get a suborder isomorphic to $\mathbb{R}$. To ensure that what we end up with is actually a subgroup, we build it as a limit of subspaces within $S$ as one would do within $\mathbb{R}$ itself when searching for a basis for $\mathbb{R}$ over $\mathbb{Q}$. At every stage we choose some linearly independent element to fill a new cut in $Q$, never choosing more than one point from a single cut, and then close to form a new subspace.

Explicitly, we do an induction of length continuum. At stage $\alpha$, choose some $s_{\alpha} \in S$ to be any element in some as yet unfilled cut in $Q$, such that $s_{\alpha} \not \in Q_{\alpha}$, where $Q_{\alpha}$ denotes the subspace of $S$ generated by $Q \cup \{s_{\beta}: \, \beta <\alpha\}$. There is some element $r_{\alpha}$ in $\mathbb{R}$ that sits in the corresponding cut in $\mathbb{Q}$, and this element is also linearly independent over the corresponding subspace $\mathbb{Q}_{\alpha} \subset \mathbb{R}$. After continuum many steps we will have filled all gaps in $Q$, and constructed the subspace $R \subset S$ generated by $Q \cup \{s_{\alpha}: \alpha < 2^{\aleph_0} \}$. This space is necessarily isomorphic to $\mathbb{R}$ as an additive group. To see this, note every point in $R$ has a unique representation as $q_01 + q_1s_{\beta_1} + \ldots + q_ks_{\beta_k}$, where the $q_i \in \mathbb{Q}$. Then the isomorphism is the obvious one: just extend the map that takes each $s_{\alpha}$ to $r_{\alpha}$.

It remains to verify that $S$ is a vector space over $\mathbb{Q}$. We'll sketch how to show that $\frac{1}{2}x$ exists for ever $x \in S$. Assume for simplicity that $x$ is positive. Proceed as one would guess: pick some $x_0$ between $0$ and $x$. Then $x_0 + x_0$ is either equal to, less than, or greater than $x$. In the first case we are done; in the second, pick an $x_1$ between $x_0$ and $x$; in the third, pick $x_1$ to be between $0$ and $x_0$. And so on, in this way building two sequences, one increasing and one decreasing (and one possibly empty), that seek to converge to $\frac{1}{2}x$. At successor stages go as before, and at limit stages take limit points, which always exist as $S$ is complete. This process will terminate, in the sense that either at some stage we find $\frac{1}{2}x$, or we keep going until our upper sequence converges to some $y \leq \frac{1}{2}x$, or the lower sequence converges to some $y \geq \frac{1}{2}x$. There are some details to check, but one can show that in each case the point to which the relevant sequence converges must be $\frac{1}{2}x$.

Analogously, one may find $\frac{1}{n}x$, and so $\frac{m}{n}x$. From there it is not too difficult to show that these rational scalars satisfy the vector space axioms. We are done.

It's worth noting that we use in an essential way the ccc-ness and completeness of Suslin lines to get a contradiction. Nothing like this argument can be used in the case of Aronszajn lines, and it seems plausible to me that such a line could be the order type of an abelian group.