Incnis Mrsi's nice argument from 2014 has not actually been completed so let's complete it, since this came up recently, twice (actually a third time on MO, too).
We have a topological group $G$ with identity $e$ homeomorphic to $\mathbb{R}$. Incnis has established that this group is totally ordered with respect to the usual order on $\mathbb{R}$ (in a way compatible with the topology; that is, $G$ is totally ordered, the total order is isomorphic to $\mathbb{R}$, and the isomorphism is a homeomorphism also), and in particular torsion-free. Thanks to the total order it follows that for any positive integer $n$, the $n^{th}$ power function $g \mapsto g^n$ is strictly monotonic and hence injective.
Proposition: $g \mapsto g^n$ is unbounded from above and below.
Proof. If the image of $g \mapsto g^n$ contains some positive $L > e$ then it also contains $L^2 > L$, which is strictly larger. Similarly for negatives. $\Box$
Corollary: The function $g \mapsto g^n$ is bijective; equivalently, every element has a unique $n^{th}$ root.
Proof. By the above, the image of $g \mapsto g^n$ contains $e$ and is unbounded from above and below. Since it is also connected, the image is all of $G$, so $g \mapsto g^n$ is surjective. Since it is also strictly monotonic, it is injective (as observed above), hence bijective. $\Box$
Pick any $g > e$ in $G$ and define $f : \mathbb{Q} \to G$ by sending $\frac{p}{q}$ to the unique $q^{th}$ root of $g^p$. This is well-defined because if $\frac{p}{q} = \frac{r}{s}$ then $g^{ps} = g^{qr}$ has a unique $qs$-th root. It is a homomorphism for the same reason. And it is strictly monotonic, hence extends continuously to a homomorphism $F : \mathbb{R} \to G$ by a limiting argument. (This should all be very familiar if you've ever constructed the exponential $a^x$ in terms of limits of rational exponentials in a real analysis class, and in fact that argument is a special case of this one, applied to $G = (\mathbb{R}_{+}, \times)$.)
$F$ is strictly monotonic, hence injective. We will write $F(r) = g^r$.
Proposition: $r \mapsto g^r$ is unbounded from above and below.
Proof. If the image of $r \mapsto g^r$ contains any positive $L > e$ then it also contains $gL > L$, which is strictly larger. Similarly for negatives. $\Box$
Corollary: $r \mapsto g^r$ is surjective, hence an isomorphism of topological groups $G \cong (\mathbb{R}, +)$.
Proof. The image of $r \mapsto g^r$ is unbounded from above and below, contains $e$, and is connected, hence is all of $G$. So $r \mapsto g^r$ is surjective, hence a bijection. Since it's strictly increasing (hence continuous), its inverse is also strictly increasing (hence continuous), so it's a homeomorphism. $\Box$
Corollary: $G$ is abelian.