The most general situation I can cover is the following:
Proposition. Let $k$ be a field of characteristic $0$, and let $R$ be a $k$-algebra of finite type. If $f,g \in R[x,y]$ are polynomials such that $\det\operatorname{Jac}(f,g) = 1$ but $(f,g) \colon R[x,y] \to R[x,y]$ is not an isomorphism, then $\deg f \geq 100$ or $\deg g \geq 100$.
The proof relies on the following lemma:
Lemma. Let $S$ be a scheme, let $X$ and $Y$ be flat $S$-schemes that are locally of finite presentation, and let $f \colon X \to Y$ be a morphism of $S$-schemes. If $f_s \colon X_s \to Y_s$ is an isomorphism for all $s \in S$, then $f$ is an isomorphism.
Proof. The assumptions imply $f$ is flat by the fibrewise criterion of flatness [Tag 039E]. Hence, $f$ is universally open [Tag 01UA]. Since each fibre $X_y \to y$ is an isomorphism, we conclude that $f$ is a universal homeomorphism, so in particular $f$ is affine [Tag 04DE]. It also follows that $f$ is universally closed, hence proper. Thus, $f$ is finite [Tag 01WN], hence finite locally free of rank $1$. It is now an easy exercise (cf. e.g. this blog post I wrote) that this forces $f$ to be an isomorphism. $\square$
Remark. Note that there are easy counterexamples without the flatness assumptions. For example, we can take $S = Y$ to be a nodal curve, and $X$ its normalisation minus one of the points above the node. This should convince you that the result is less trivial than it sounds.
Proof of Proposition. By the lemma, if $\phi = (f,g) \colon R[x,y] \to R[x,y]$ is not an isomorphism, then the same must be true for $\phi \otimes \kappa(\mathfrak p) \colon \kappa(\mathfrak p)[x,y] \to \kappa(\mathfrak p)[x,y]$ for some prime ideal $\mathfrak p \subseteq R$. Since the map $\kappa(\mathfrak p) \to \overline{\kappa(\mathfrak p)}$ is faithfully flat, we conclude that $\phi \otimes \overline{\kappa(\mathfrak p)}$ is also not an isomorphism.
But the condition $\det \operatorname{Jac}(f,g) = 1$ propagates to their images $\bar f, \bar g \in \overline{\kappa(\mathfrak p)}[x,y]$, so Moh's result shows that $\deg \bar f \geq 100$ or $\deg \bar g \geq 100$. This forces $\deg f \geq 100$ or $\deg g \geq 100$. $\square$