It is not difficult to check that the set of continuous functions from $\mathbb{R}$ to $\mathbb{R}$ is a ring (an $\mathbb{R}$-algebra), and similarly (if I am not wrong), the set of continuous functions from $[a,b]$ to $\mathbb{R}$ is a ring (an $\mathbb{R}$-algebra), where $a,b \in \mathbb{R}$, $a < b$, denote it by $R_{[a,b]}$.
Let $R_{[0,1]}[X,Y]$ be the polynomial ring in $X$ and $Y$ over $R_{[0,1]}$.
Denote, for all $t \in [0,1]$: $f(t)=\sqrt{t}$, $g(t)=t$, $h(t)=1$, and let $A=fX+gY$, $W=hX+fY$
We have, $\operatorname{Jac}(A,W):=A_xW_y-A_yW_x=f^2-gh=t-t=0$.
Here, $W \notin R_{[0,1]}[A]$, since $W=hX+fY=\frac{1}{f}(fX+gY)=\frac{1}{f}A \notin R[A]$ (since $\frac{1}{f}(t)=\frac{1}{\sqrt{t}}$ is not defined for $0$).
However, $A$ does not have a Jacobian mate in $R_{[0,1]}[X,Y]$, since for every $B \in R_{[0,1]}[X,Y]$: $\operatorname{Jac}(A,B)=fB_y-gB_x=fB_y-f^2B_x=f(B_y-fB_x) \notin R_{[0,1]}^{\times}$.
First question: Does there exist $A,B,W \in R_{[0,1]}[X,Y]$ satisfting the following 3 conditions:
(1) $\operatorname{Jac}(A,W)=0$.
(2) $W \notin R_{[0,1]}[A]$.
(3) $\operatorname{Jac}(A,B) \in R_{[0,1]}^{\times}$.
Perhaps Weierstrass function may help? Perhaps trigonometric functions may help? It seems plausible that $A$ may not be linear, namely, $A$ will have total degree $>1$.
Second question: What if we consider complex (analytic) functions instead of real continuous functions?
Third question, general case: Replace those rings of functions by other rings.
The general case has one plausible easy answer, thanks to Warning 1.1.17, adjusted to my question: Let $R:=\frac{\mathbb{Z}[t]}{(2t)}$, $R[X,Y]$, $A=X-\bar{t}X^2$, $B=Y$, $W=X$. We have:
(1) $\operatorname{Jac}(A,W)=(1-2\bar{t}X)0-01=10-01=0$.
(2) $W=X \notin R[X-\bar{t}X^2]$ from considerations of degrees.
(3) $\operatorname{Jac}(A,B)=1$.
Therefore, in the general case we should add a fourth condition:
(4) $R$ is a $\mathbb{Q}$-algebra.
Any hints and comments are welcome!