Do all fields with internal absolute values arise either as ordered fields or like $\mathbb{C}$ from them? (I have now also asked this on mathoverflow.)

Let ​ $\langle F,\hspace{-0.04 in}+,\hspace{-0.03 in}\cdot \rangle$ ​ be a field, let $E$ be a non-zero strict subring of $F$, let ​ $\leq$ ​ be

a total order on $E$ that makes ​ $\langle E,\hspace{-0.04 in}+,\hspace{-0.03 in}\cdot,\hspace{-0.03 in}\leq \rangle$ ​ into an ordered ring, and let

$|\hspace{-0.05 in}\cdot \hspace{-0.05 in}| : F\to E$ ​ be such that for all elements $x$ of $E$, for all elements $y$ and $z$ of $F\hspace{-0.02 in}$,
$0\leq x \; \implies \; |x| = x$
and
$0\leq |z| \;\;\;\;$ and $\;\;\;\; |\hspace{.02 in}y\hspace{-0.04 in}\cdot \hspace{-0.04 in}z| \; = \; |y|\cdot |z| \;\;\;\;$ and $\;\;\;\; |\hspace{.02 in}y\hspace{-0.03 in}+\hspace{-0.03 in}z| \; \leq \; |y|+|z|$
.

Does it follow that there exists an element $i$ of $F$ such that
  
  $i^{\hspace{.03 in}2} = -1$
  and
  for all elements $z$ of $F$, there exist elements $a$ and $b$ of $E$
  
  such that $\;\;\;\;\;\;\; z \; = \; a\hspace{-0.03 in}+\hspace{-0.03 in}(b\hspace{-0.05 in}\cdot \hspace{-0.05 in}i\hspace{.02 in}) \;\;\;$ and $\;\;\; |z|^{\hspace{.02 in}2} \; = \; a^{\hspace{.02 in}2} + b^{\hspace{.02 in}2}$
  
  ?


One can verify that the implication and the multiplicativity

force $E$ to be a sub$field$ of $F$ and force $|\hspace{-0.05 in}\cdot \hspace{-0.05 in}|$ to satisfy

for all non-zero elements $y$ of $F$, $\;\;\; |\hspace{.02 in}y| \neq 0 \;$ and $\; \left|\hspace{.03 in}y^{-1}\right| = |\hspace{.02 in}y|^{-1} \;\;\;$.
Accordingly, a necessary condition for $i$ to satisfy the conditions of my question

is that ​ $\{\hspace{-0.03 in}1,\hspace{-0.03 in}i\}$ ​ be a basis for $F$ as a vector space over $E$.

I now see how to prove that letting $F$ be a subfield of $\mathbb{C}$ and letting $|\hspace{-0.05 in}\cdot \hspace{-0.05 in}|$

agree with the usual absolute value on $\mathbb{C}$ cannot yield a counterexample.


Lemma: ​ For all such structures, all real elements of $F$ are in $E$.
Proof: ​ ​ ​ ​ ​ ​ ​ Fix any such structure and let $z$ be any real element of $F$. ​ ​ ​ $|z| = \pm z$ ​ and ​ $|z|\in E$ ,

so at least one of ​ $\{z\hspace{.02 in},\hspace{-0.04 in}-z\}$ ​ is in $E$. ​ ​ ​ Since $E$ is a subring, that means ​ $z\in E$ .

Result: ​ Such structures cannot be counterexamples.

Proof:
Let $z$ be any element of $F$ that's not in $E$. ​ ​ ​ By the contrapositive of the lemma, ​ $|z|\not\in \mathbb{R}$ .

In other words, ​ $\operatorname{Im}(z) \neq 0$ , ​ so in particular ​ $z\neq 0$ .

$|z|^2 \; = \; \overline{z} \cdot z \qquad \qquad \overline{z} \; = \; |z|\hspace{-0.03 in}\cdot \hspace{-0.03 in}|z|\hspace{.04 in}/\hspace{.04 in}z \; \in \; F \qquad \qquad 2\cdot \operatorname{Im}(z) \cdot i \; = \; \overline{z}-z \; \in \; F$

$\operatorname{Im}(z) \cdot i \: = \: 2\hspace{-0.03 in}\cdot \hspace{-0.03 in}\operatorname{Im}(z)\hspace{-0.03 in}\cdot \hspace{-0.03 in}i\hspace{.04 in}/\hspace{.04 in}2 \: \in \: F \quad \quad \quad i \: = \: \operatorname{Im}(z)\hspace{-0.03 in}\cdot \hspace{-0.03 in}i\hspace{.04 in}/\operatorname{Im}(z) \: = \: \operatorname{Im}(z)\hspace{-0.03 in}\cdot \hspace{-0.03 in}i\hspace{.06 in}/\hspace{.04 in}|\hspace{-0.04 in}\operatorname{Im}(z)\hspace{-0.03 in}\cdot \hspace{-0.03 in}i| \: \in \: F$

$\operatorname{Im}(z) \; = \; \operatorname{Im}(z)\hspace{-0.03 in}\cdot \hspace{-0.03 in}i\hspace{.04 in}/\hspace{.04 in}i \; \in \; F \qquad \qquad \operatorname{Re}(z) \; = \; z-(\operatorname{Im}(z)\hspace{-0.03 in}\cdot \hspace{-0.03 in}i) \; \in \; F$

By the lemma, $\operatorname{Re}(z)$ and $\operatorname{Im}(z)$ are both in $E$.

 A: $\DeclareMathOperator{\id}{id}$
I am editing my message yet again to bring you if not an answer, a few ideas.
What I was trying to find as a counter example was an ordered field $F$ equipped with its absolute value $| \ |_F$, a proper subfield $E$ and a morphism $\varphi: F_{>0} \rightarrow E_{>0}$ fixing $E_{>0}$ (we would then define $\varphi(0) := 0$ and $\forall x \in F_{>0}, \varphi(-x):= -\varphi(x)$) satisfying $\forall x,y \in F^{\times}, |\varphi(x+y)|_F \leq |\varphi(x)|_F + |\varphi(y)|_F$. The reason I thought this probably existed is that if $E^{\times},F^{\times}$ being abelian groups without torsion, there are plenty of morphisms $F_{>0} \rightarrow E_{>0}$ defined by choosing a rationally independant basis* of $E_{>0}$ and completing it into a rationally independant basis of $F_{>0}$, then taking projections or other sorts of endomorphisms. However, this does not insure the condition $\forall x,y \in F^{\times}, |\varphi(x+y)|_F \leq |\varphi(x)|_F + |\varphi(y)|_F$, and we even know that this condition can never hold in some cases, for instance if $E$ is dense in $F$, because it forces $\varphi$ to be continuous with respect to the order topology, and so $\varphi$ should be $\id_F$. 
So I tried findind concrete examples rather than choice-based ones and I couldn't. It would be interesting to understand whether this type of method can work or not.
*it's like a basis over $\mathbb{Q}$ (every element is a unique linear combination in this basis with coefficients in $\mathbb{Q})$) except not every linear $\mathbb{Q}$-combination need have a meaning in the group because not every element need have a $n$-th root for every $n$. It exists (and can be completed like a regular basis can) for any abelian group without torsion, assuming AC.
