Let $f : (x,y) \mapsto (p,q)$ be a map from $\mathbb{C}[x,y]$ to $\mathbb{C}[x,x^{-1},y]$ satisfying the following two conditions:
(i) $\operatorname{Jac}(p,q):=p_xq_y-p_yq_x \in \mathbb{C}-\{0\}$.
(ii) $l_{1,-1}(p)=\lambda x^2y$ and $l_{1,-1}(q)=\mu x^{-1}$, for some $\lambda,\mu \in \mathbb{C}-\{0\}$. Then we can write $p=\lambda x^2y + A$ and $q=\mu x^{-1}+B$, where $v_{1,-1}(A)<1$ and $v_{1,-1}(B) < -1$.
Is it true that (i)+(ii) imply that $A=B=0$?
In other words, is it true that (i)+(ii) imply that $p=\lambda x^2y$ and $q=\mu x^{-1}$?
Otherwise, could one find $f$ satisfying (i)+(ii) having $A \neq 0$ or $B \neq 0$?
Edit: Variations and generalizations on the above question:
(1) Taking $f: (x,y) \mapsto (\lambda x^{-1}y+A,\mu x^2+B)$.
(2) More generally, taking $f: (x,y) \mapsto (\lambda x^{n+1}y+A,\mu x^{-n}+B)$, $0 \neq n \in \mathbb{Z}$.
(3) Replacing $k[x,y] \to k[x,x^{-1},y]$ by $A_1(k) \to A_1(k)^{-1}$, in all questions, especially the general case (2).
Where $A_1(k)$ is the first Weyl algebra, namely, the associative non-commutative $k$-algebra generated by $x$ and $y$ subject to the relation $yx-xy=1$. And $A_1(k)^{-1}$ is the ring extension of $A_1(k)$ generated by $x,x^{-1},y$ subject to $yx-xy=1$, $xx^{-1}=x^{-1}x=1$ and $[y,x^{-1}]=-x^{-2}$.
(4) Considering $k[x^{1/r},x^{-1/r},y]$ instead of $k[x,x^{-1},y]$, $r \in \mathbb{N}$. And considering $A_1(k)^{-1/r}$ instead of $A_1(k)^{-1}$.
Thank you veru much! I now asked the above question in MO.