Let ${(A, P)}$ be a preordered $\mathbb{R}$-algebra in the sense that $A$ is a $\mathbb{R}$-algebra and ${P \subseteq A}$ is a subset closed under addition, multiplication, containing the nonnegative reals and not containing ${-1}$.
Assume also that we are given an ${x, y \in P}$ and a morphism ${f : (R, P) \rightarrow (\mathbb{R}, \mathbb{R}_{\geq 0})}$ of preordered algebras such that ${f x < f y}$.
Question: Can there exist a ${g : (R, P) \rightarrow (\mathbb{R}, \mathbb{R}_{\geq 0})}$ such that ${g y < g x}$?
Note: I have the impression that under reasonable not too strong hypotheses such a $g$ should not exist. So I guess that what I am really asking is for the weakest hypotheses that guarantee that no such $g$ exists. For instance, is lack of direct product decompositions enough? (Of course, requiring $P$ to be totally ordered would be pointless.) Anyway, I don't know if my intuition is correct.
How about the following: let $A=\Bbb R[x,y]$ with $P$ the set of polynomials with all coefficients non-negative. Then $x,y\in P$, and the the morphism $f$ given by evaluating $x=1$ and $y=2$ gives $fx<fy$, but the morphism $g$ given by evaluating $x=2$ and $y=1$ gives $gy<gx$.
I think your intuition could be good if one requires some sort of comparability between $x$ and $y$: for instance, if $y-x \in P$, then you get what you want (though this is maybe too trivial?).