Given that $E$ is a linearly ordered set, such that $E$ has no maximal element.
Then we want to consider the set of formal products of elements of $E$ (i.e. elements of the form $e_1 * e_2 * ... * e_n$ such that $e_1, e_2, ..., e_n \in E$). To do this (using my personal approach), I consider the set: $$E^{(1)} = \bigcup_{n \in \mathbb{N}} E^n$$
(where $E^n = \{$functions from $n$ to $E \}$).
In order to account for repeated permutations, for each $f \in E^{(1)}$ define the function $\#_f \colon E \to \mathbb{N}$ to be given by $\#_f(e) = \mathrm{Card}(\{n \in \mathbb{N} : f(n) = e\})$. Then consider the equivalence relation $\sim$ on $E^{(1)}$ given by $f \sim g$ if and only if $\#_f = \#_g$, so that we can take the quotient $E^{(2)} = E^{(1)} / \sim$.
Then we define the linear order on $E^{(2)}$ to be given by the following equivalence:
- $[f]_\sim \leq [g]_\sim$
- For all $e \in E$, either $\#_f(e) \leq \#_g(e)$ or there exists $\epsilon \in E$ with $e < \epsilon$ and $\#_f(\epsilon) < \#_g(\epsilon)$.
This gives us a well-defined linear order on the set of formal products of $E$ (where all elements of $E$ are treated as positive, and the empty product represents $1$). We quickly add in negative elements by taking a copy of $E^{(2)}$ with reverse ordering (which we can denote by $\mathrm{Neg}(E^{(2)})$), and define $E^{(3)} = \mathrm{Neg}(E^{(2)}) \sqcup E^{(2)}$ with linear order given by having all the elements of $\mathrm{Neg}(E^{(2)})$ precede all the elements of $E^{(2)}$.
Given this ordering, there is a canonical way to extend our linear order to the set of formal sums of $E^{(3)}$ (i.e. elements of the form $e_{1,1}*...*e_{1,n_1} + ... + e_{m,1}*...*e_{m,n_m}$), such that if $x, y \in E^{(3)}$ with $x < y$ and $n \in \mathbb{N} \backslash \{0\}$ then $n*x < y$. The details are left out due to being tedious, but can be added if requested.
Which leaves us with an ordered ring $\mathrm{Ring}(E)$, that in turn gives us an ordered field $\mathrm{Field}(E)$ (the field of fractions of $\mathrm{Ring}(E)$).
Note that $E$ is cofinal in $\mathrm{Field}(E)$. So if $E$ has a well-ordered cofinal subset, then $\mathrm{Field}(E)$ has a well-ordered cofinal subset.
If $\mathrm{Field}(E)$ has a well-ordered cofinal subset (say $S$). Without loss of generality, we can assume that every element of $S$ is positive. Then there exists a unique ordinal $\alpha$ and a unique order isomorphism $h \colon \alpha \to S$.
- Then for all $s \in S$ there exists unique $p_s, q_s \in \mathrm{Ring}(E)$ with $s = \frac{p_s}{q_s}$, such that $\frac{p_s}{q_s}$ is in reduced form.
- Then $P = \{ p_s \in \mathrm{Ring}(E) : s \in S \}$ is a cofinal subset of $\mathrm{Field}(E)$ as $p_s \geq s$ for all $s \in S$, due to $1$ being the smallest positive element in $\mathrm{Ring}(E)$. Further we have that $h^{*} \colon \alpha \to P$ given by $h^{*}(a) = p_{h(a)}$ is a well-defined surjective function.
- Then for all $p \in P$ there exists a unique $t_p \in E^{(2)}$ such that $t_p$ is the leading term in $p$, as every element of $P$ is a unique sum of finitely many elements in $E^{(2)}$.
- Then $T = \{ t_p \in E^{(2)} : p \in P \}$ is a cofinal subset of $\mathrm{Field}(E)$, as $E$ has no maximal element. Further we have that $h^{**} \colon \alpha \to T$ given by $h^{**}(a) = t_{h^{*}(a)}$ is a well-defined surjective function.
- Then for all $t \in T$ there exists a unique $e_t \in E$ such that $e_t$ is the maximal value in the product, as every element of $T$ is a unique product of finitely many elements of $E$.
- Then $X = \{ e_t \in E : t \in T \}$ is a cofinal subset of $\mathrm{Field}(E)$, as $E$ has no maximal element. Further we have that $h^{***} \colon \alpha \to X$ given by $h^{***}(a) = e_{h^{**}(a)}$ is a well-defined surjective function.
- Then $X$ is a cofinal subset of $E$, as $E$ is cofinal in $\mathrm{Field}(E)$.
- Then we can consider the subset $Y = \{ h^{***}(a) \in X : \forall b \in a, h^{***}(b) < h^{***}(a) \}$, which is cofinal in $E$ and well-ordered by $h^{***}|_Y$ with the order of $h^{***}|_Y$ agreeing with the order of $E$.
- Then $Y$ is a well-ordered cofinal subset of $E$.
Then $E$ has a well-ordered cofinal subset, if and only if, $\mathrm{Field}(E)$ has a well-ordered cofinal subset.
So if every ordered field has a well-ordered cofinal subset, then every linearly ordered set without a maximal element has a well-ordered cofinal subset (as $E$ is arbitrary).
So if every ordered field has a well-ordered cofinal subset, then every linearly ordered set has a well-ordered cofinal subset (as any linearly ordered set with a maximal element clearly has a well-ordered cofinal subset).
Hence every ordered field has a well-ordered cofinal subset, if and only if, every linearly ordered set has a well-ordered cofinal seubset (as the reverse claim is trivial).