Let $k$ be a field of characteristic zero, and let $R$ be a commutative $k$-algebra, which is perhaps non-affine and/or not an integral domain. For example: $R=\frac{\mathbb{Q}[x_1,x_2,\dots]}{(x_1x_2)}$.
Is it possible to find $A,B,W \in R[X,Y]$ such that: (i) $\operatorname{Jac}(A,B)=1$, (ii) $\operatorname{Jac}(A,W)=0$, (iii) $W \notin R[A]$?
I only succeeded to find elements $A,B,W \in R[X,Y]$ satisfying (i) + (iii) or (ii) + (iii), but not satisfying the three conditions.
Several attempts:
$R=k[t^2,t^3]$, $A=t^2X+t^3Y$, $W=t^3X+t^4Y$ satisfy (ii)+(iii). Indeed, (ii) is clear, $W=tA \in k(t^2,t^3)[A]-k[t^2,t^3][A]$, and (i) is not satisfied (since $\operatorname{Jac}(A,B)=t^2 E$, for some $E \in R[X,Y]$, never equals $1$).
Taking $R=\frac{k[t^2,t^3,s]}{(t^2s=1)}$ does not help, since now $R \ni t^3s=t(t^2s)=t1=t$, so although $A$ has a Jacobian mate, $B=sY$ (=(i) is satisfied), now (iii) is not satisfied, since $W=tA \in R[A]$.
Taking $R=k((t))$ also does not help. Now $1-t^2$ is invertible, with inverse $\epsilon:=1+t^2+t^4+t^6+\dots$. $A=(1-t^2)X+(1-t)Y$ has a Jacobian mate $B=\epsilon Y$. $W=(1+t)X+Y$ satisfies $\operatorname{Jac}(A,W)=0$, but $W=(1-t)^{-1}A \in R[A]$, so (iii) is not satisfied.
Remark: Relying on several considerations that I will not explaim now (perhaps if someone will ask me to explain, then I will explain), it seems that $R$ should be non-affine+not an integral domain, in order to have a chance to get an example with $\deg(A),\deg(B) < 100$. In other words, it seems that if $R$ is affine or an integral domain, then each of possible $A$ and $B$ must be of degrees $ \geq 100$. Therefore, my above attempts ($R$ is affine) are hopeless. So we should better consider something like $R=\frac{k[[x_1,x_2,\dots]]}{(x_1^2)}$. $A=X+(x_1+x_2+\dots)Y$ has a Jacobian mate $B=X+Y$. Can we find an appropriate $W$? What about $R= \frac{k[[x_1,x_1^{-1},x_2,x_2^{-1},\dots]]}{I}$, for some appropriate ideal $I$?
Thank you very much!