Does every ordered divisible abelian group admit an expansion (and how many) to an ordered field? Let $(G,+,<)$ be an ordered divisible abelian group.

$1)$ Is it always the case that there exists a binary function $*:G\times G \rightarrow G$ such that $(G,+,*,<)$ is an ordered field?
$2)$ Given an archimedean group where the above holds (e.g. $(\mathbb{R},+,<)$), how many of these product operations are there?

Guess: In the archimedean case for each $g>0$ you can cosider $*$ in which you identify $g$ with $1$ and then $*$ is uniquely defined in the ordered divisible abeliean group $G_g=\{\frac{n}{m} g : n,m < \omega\}$ and expands this group to an ordered field and then, since $G$ is archimedean, $G_g$ is dense in $G$, and $*$ can be uniquely extended to a field product operation in $G$.
So if $G$ is archimedean the possible $*$ operations are in unique correspondence with each $g>0$.
 A: (1): No.  For instance, any non-archimedean ordered field must have infinite dimension as a vector space over $\mathbb{Q}$ (if $x$ is an infinitely large element, then it is transcendental over $\mathbb{Q}$).  So any non-archimedean ordering of $\mathbb{Q}^n$ for finite $n>1$ (e.g., the lexicographic order) cannot be made into an ordered field.  A bit more generally, if $G$ is non-archimedean and $x\in G$ is any positive element, then in order for $G$ to have an ordered field structure there must be an element $y$ which is greater than any integer multiple of $x$ (namely, $x^2$ if $x$ is infinitely large and any infinitely large element otherwise).
(2): Your argument is correct.  More generally, if $G$ is any archimedean divisible ordered abelian group and you fix any positive $g\in G$, there is a unique order-preserving homomorphism $i_g:G\to\mathbb{R}$ sending $g$ to $1$.  We then see that $G$ admits an ordered field structure with $g$ as the unit iff the image of this $i_g$ is a subfield of $\mathbb{R}$ (and in this case the ordered field structure is unique given $g$).  Indeed, if $G$ has an ordered field structure with $g$ as the unit then $i_g$ is automatically a field homomorphism so its image is a subfield (and the field structure must be the one on the image of $i_g$).  Conversely, if the image of $i_g$ is a subfield, you can just pull the field structure on $i_g(G)$ back to $G$.
Now if $h$ is any other positive element of $g$, then $i_h$ and $i_g$ are related by $i_h(x)=i_g(x)/i_g(h)$.  In particular, if the image of $i_g$ is a subfield, then the image of $i_g$ is the same as the image of $i_h$ (since the image of $i_g$ is closed under multiplication and division by $i_g(h)$).  So whether the image of $i_g$ is a subfield does not actually depend on the choice of $g$.
This gives many archimedean counterexamples to (1).  If $G$ is any $\mathbb{Q}$-vector subspace of $\mathbb{R}$ which contains $1$ and is not a subfield of $\mathbb{R}$, then this shows that $G$ cannot be made into an ordered field.
