The following is a special case of my original question (perhaps it is better to first concentrate on this special case):
Let $k$ be a field of characteristic zero, $m \in \mathbb{Z}$, and $f: (x,y) \mapsto (p:=x^{1+m}y+a,q:=x^{-m}+b)$, a $k$-algebra homomorphism from $k[x,y]$ to $k[x,x^{-1/m},y]$ having a non-zero scalar Jacobian, namely, $\operatorname{Jac}(p,q):=p_xq_y-p_yq_x \in k-\{0\}$.
Here I assume that $l_{1,-1}(p)=x^{1+m}y$ and $l_{1,-1}(q)=x^{-m}$, hence $\deg_{1,-1}(a)<m$ and $\deg_{1,-1}(b)<-m$.
Claim: Such $f$ necessarily satisfies: $a \in k[x,x^{-1}]$ (of appropriate $(1,-1)$-degree) and $b=0$.
I have a sketch of proof of that claim, based on considerations of $(1,-1)$-degrees and the Jacobians of the $(1,-1)$-homogeneous components of $a$ and $b$.
Is the claim true, or can one find a counterexample?
Notice that $pq=xy+E$, where $\deg_{1,-1}(E) < 0$; perhaps this information may help?
Please also see this question.
Any hints and comments are welcome! Thank you.
You have to use the language and results from Gucc. et al If $m>0$, then it suffices to prove that $\min\{Succ_P(1,-1),Succ_Q(1,-1)\}>(0,1)$ (See Definition 3.4 of Gucc. et al for the definition of $Succ$).
Now assume by contradiction that $$ (0,1)\ge (\rho,\sigma) :=\min\{Succ_P(1,-1),Succ_Q(1,-1)\}>(1,-1). $$ Then $v_{\rho,\sigma}(p)>0\ge v_{\rho,\sigma}(q)$, and it follows that $en_{\rho,\sigma}(p)\nsim en_{\rho,\sigma}(q)$. By Proposition 2.4 of Gucc. et al we have $$ (0,0)=en_{\rho,\sigma}([p,q])=en_{\rho,\sigma}(p)+en_{\rho,\sigma}(q)-(1,1). $$ Hence the only possibility is $en_{\rho,\sigma}(p)=(1+m,1)$ and $en_{\rho,\sigma}(q)=(0,-m)$, which contradicts the fact that either $\mathcal{l}_{\rho,\sigma}(p)$ or $\mathcal{l}_{\rho,\sigma}(q)$ is not a monomial (else $(\rho,\sigma)$ wouldn't be $Succ_P$ nor $Succ_Q$.)
If $m<0$, then there are counterexamples, for example, if $m=-1$, then $p=y+\frac{3y^4}{x^2}$ and $q=x+\frac{12y^3}{x}+\frac{24y^6}{x^3}$ gives a counterexample.
For $m=-2$ we can take $p=x^{−1}y+3x^{−8}y^4$ and $q=x^2+12x^{−5}y^3+24x^{−12}y^6$.
For general $m$ we can take $p=y x^{1+m}+3y^4 x^{4+6 m}$ and $q=x^{-m}+12 y^3 x^{3+4 m}+24y^6 x^{6+9 m}$
For the extended Weyl algebra the proof for $m>0$ stands its ground, but the construction of counterexamples is a little bit trickier.