What do you get if you perform the Dedekind Cuts procedure on $\mathbb Q(x)$? Let $\mathbb Q(x)$ denote, as usual, the field of rational functions with rational coefficients.  Any element of $\mathbb Q(x)$ can be written in the form
$$f=\frac{a_n x^n + \cdots +a_0}{b_m x^m + \cdots +b_0}$$
with $a_n, b_m \ne 0$ and where all coefficients are integers. Let $P \subset F$ be defined by the condition $f\in P \iff \frac{a_n}{b_m}> 0$, and define $f \prec g \iff g-f \in P$.  Then $(F, \prec)$ is an ordered field.
As is discussed in this answer, under this order $\mathbb Q(x)$ is non-Archimedean.  In particular, $x$ is a transfinite element (in the sense that $r \prec x$ for all $r \in \mathbb Q$) and $\frac 1x$ is infinitesimal (in the sense that $0 \prec \frac 1x \prec r$ for all positive $r \in \mathbb Q$).  The fact that $\mathbb Q(x)$ is non-Archimedean means it is not complete, which is easily witnessed by considering the set $\mathbb Q \subseteq \mathbb Q(x)$; this subset is bounded (for example, by $x$) but has no least upper bound.
What happens if we now perform the Dedekind cut construction on $\mathbb Q(x)$?  What do we get?
Any complete ordered field is isomorphic to $\mathbb R$, and certainly the Dedekind completion of $\mathbb Q(x)$ is larger than $\mathbb R$, so this must not be a complete ordered field.  But why not?  Is the result no longer a field?  Or do we end up with a field for which the axioms for an ordered field are somehow broken?
Related question here, but that question starts with the hyperreals, which (I think?) is not the same question as mine.  (It's possible that the answer to my question is "You get the completion of the hyperreals"; if so, I'd love to understand that better!)
 A: If we define $\deg f=n-m.$ Then if $f,g\in P,$ and $\deg f<\deg g,$ then $f\prec g.$ Define $P_d=\{f\in P\mid \deg f=d\}.$
Define $\beta=\inf P_1.$ This would be the Dedekind cut $\{f\in P\mid \deg f<1.\}$
Now, given any $f\in P,$ $f\in P_1$ iff $f+1\in P_1.$
So, if the result of the Dedekind process is a field, we'd get $\beta+1=\beta,$ or $1=0.$
So there must be a problem with the definition of addition - that it no longer has the property that if $\alpha\neq 0$ then $\alpha+\beta\neq \beta.$
I'm still thinking about what specific property of $\mathbb Q$ lets us get this property, while this field does not. But I think it is related to the fact that, for rationals $q,r>0$ then $q+q+q+q+\cdots + q>r$ for some number finite sum of values $q.$ This is essentially the Archimedean property on the rational numbers.

The rational numbers have the property that in a nonzero cut, $\alpha,$ there is a sequence $(m_k,M_k)$ with $m_k\in\alpha, M_k\notin\alpha,$ and $\inf_k (M_k-m_k)=0.$
This is usually constructed by starting with any pair $(m_0,M_0)$ and then recursively computing $m=(m_k+M_k)/2,$ and replacing either $m_k$ or $M_k$ with $m,$ depending on whether $m\in\alpha$ or not.
Given a positive $p\in\beta,$ then, there is some $k$ such that $p>M_k-m_k$ and thus $p+m_k\in\alpha+\beta,$ but $p+m_k\notin \alpha.$
But in your totally ordered abelian group, we cannot deduce that $\inf (M_k-m_k)=0.$ Indeed, $\deg (M_k-m_k)=\deg(M_0-m_0)=d,$ and thus $x^{d-1}\in P$ is a positive lower bound for $\{M_k-m_k\}.$
And, indeed, if $m_0\in P_0,M_0\in P_1,$ we'd get all $m_k=m_0$ and all $M_k\in P_1.$ So you can deduce that $M_{k}-m_0>m_0$ for all $k.$
