The germ of Tarski's construction of multiplication is the observation that the structure $\mathcal{R}_0=(\mathbb{R}; +, 1,<)$ is rigid (= has no nontrivial automorphisms). This gives us hope that multiplication is "definable" in $\mathcal{R}_0$ in some reasonable sense (although perhaps not via a first-order formula).
Moving to the "nonstandard" setting, our very first question should then be:
Is there a non-Archimedean field $\mathcal{F}=(F; +,\times,0,1,<)$ such that every automorphism of $\mathcal{F}_0=(F; +,1,<)$ be an automorphism of $\mathcal{F}$ itself?
The answer turns out to be no, and I think this is a strong negative result relevant to your question. Interestingly, however, this doesn't seem to be trivial! Here's the proof I see at the moment:
First, we need to give $\mathcal{F}$ a vector space structure over a "sufficiently rich" field. Let $\mathbb{A}$ be the poset of all Archimedean subfields of $\mathcal{F}$, ordered by inclusion. By Zorn's Lemma this has a maximal element; fix such a $K$.
(I recall there being a name for maximal Archimedean subfields of non-Archimedean fields, but I can't track it down at the moment; if I can find a reference, I'll add it later.)
$\mathcal{F}$ has a natural structure as a $K$-vector space (more generally, given any field inclusion $H_1\subseteq H_2$ there is a canonical way to view $H_2$ as a vector space over $H_1$), and as a $K$-vector space $\mathcal{F}$ has the following property:
For every finite $x\in\mathcal{F}$ there is exactly one $k\in K$ such that $x-k$ is infinitesimal.
The existence of at most one such $k$ follows since $K$ is Archimedean, while the existence of at least one such $k$ follows from the maximality of $K$ in $\mathbb{A}$. This property becomes relevant at the end of the proof: it's needed to show that a certain type of automorphism of the reduct $(F;+,1)$ can WLOG be assumed to respect the ordering $<$.
Next, we whip up a "near-miss" construction: a simple way of producing automorphisms of $(F;+,1)$ which do not respect the multiplicative structure. Showing that there exist automorphisms produced this way which also preserve the ordering will be our final step.
Fix a basis $B$ for $\mathcal{F}$ as a $K$-vector space such that $1\in B$. The map $\alpha_B:B\rightarrow F$ defined by $\alpha(b)=2b$ for $b\in B\setminus\{1\}$ and $\alpha(1)=1$ extends to an automorphism $\hat{\alpha}_B$ of $(F;0,+)$ as a $K$-vector space. Since $\hat{\alpha}_B(1)=1$, we have in fact that $\hat{\alpha}_B$ is an automorphism of $(F;+,1)$.
Now let's show that $\hat{\alpha}_B$ does not respect multiplication. Specifically, fixing $b\in B\setminus\{1\}$ let $b^2=\sum_{1\le i\le n} a_ib_i$ with $a_i\in K$, $b_i\in B$, $b_1=1$, and $b_2=b$. Note that I'm not demanding that the $a_i$s be nonzero; however, we must have $\sum_{2\le i\le n}a_ib_i\not=0$ since $b$ is nonstandard. We then have $$\hat{\alpha}_B(b^2)=a_1+2a_2b+..+2a_nb_n$$ but $$\hat{\alpha}_B(b)\hat{\alpha}_B(b)=2b\cdot 2b=4b^2=4a_1+4a_2b+...+4a_nb_n.$$ If $\hat{\alpha}_B$ is to respect multiplication, then these must be equal, which would give $$3a_1+2(a_2b+...+a_nb_n)=0.$$ The term $2(a_2b+...+a_nb_n)$ is nonzero and so nonstandard, per the observation at the end of the previous paragraph; since $3a_1$ is standard, this gives a contradiction.
Unfortunately, $\hat{\alpha}_B$ need not respect the ordering. To ensure that the ordering is respected we need a further condition on the original basis $B$:
Say that a basis $B$ for $\mathcal{F}$ as a $K$-vector space is good if $1\in B$ and no other element of $B$ is finite but non-infinitesimal.
It's easy to show that if $B$ is good then $\hat{\alpha}_B$ is order-preserving, so all that's left is to show that a good basis actually exists. It's at this point that the precise nature of $K$ is really used.
Per the property of $\mathcal{F}$ as a $K$-vector space mentioned two sections prior, let $st$ be the map sending each finite element of $\mathcal{F}$ to the unique element of $K$ to which it is infinitesimally close. Given any basis $B$ for $\mathcal{F}$ as a $K$-vector space let $$B'=\{b\in B: b=1\mbox{ or } b\mbox{ is infinite}\}\cup\{b-st(b): b\in B\mbox{ and }b\mbox{ is finite}\}.$$ It's easy to check that $B'$ is a good basis for $\mathcal{F}$ as a $K$-vector space, and so we're done.
