Let $k$ be a field of characteristic zero and let $E=E(x,y) \in k[x,y]$. Define $t_x(E)$ to be the maximum among $0$ and the $x$-degree of $E(x,0)$. Similarly, define $t_y(E)$ to be the maximum among $0$ and the $y$-degree of $E(0,y)$.
The following nice result appears in several places, for example 1 [Proposition 2.1]2 [Lemma 1.14]3 [Proposition 10.2.6]4: Let $A,B \in k[x,y]$ satisfy $\operatorname{Jac}(A,B) \in k-\{0\}$ (such $A,B$ is called a Jacobian pair). Assume that the $(1,1)$-degree of $A$, $\deg(A)$, is $>1$ and the $(1,1)$-degree of $B$, $\deg(B)$, is $>1$. Then the numbers $t_x(A),t_y(A),t_x(B),t_y(B)$ are all positive.
Is the same result holds in the first Weyl algebra over $k$, $A_1(k)$? where instead of the Jacobian we take the commutator.
Of course, we must first define $t_x(A),t_y(A),t_x(B),t_y(B)$ in $A_1(k)$; it seems to me that the same definition holds for $A_1(k)$, or am I missing something? Perhaps it is not possible to consider $E(x,0)$, where $E \in A_1(k)$?
Later, I have also asked this question in MO (in MO I have slightly elaborated on a plausible proof).
Thank you very much!
The non commutative version of Lemma 1.3 does not hold. In fact, set $A=y−x^2$ and $B=x^4/2+y^2/2−x^2y$, then $[A,B]=1$, but neither $(0,1)$ nor $(1,0)$ are in $S_B$, and $(1,0)$ is not in $S_A\cup S_B$. However, if the Dixmier Conjecture is true, then every endomorphism is an automorphism, and if for an automorphism $f(x)=A$, $f(y)=B$, and $\deg(A),\deg(B)>1$, then the numbers $t_x(A),t_y(A),t_x(B),t_y(B)$ are all positive. So a counterexample to your proposition would be a counterexample to the Dixmier Conjecture.
So your proposition is probably true, but I can't think of a way to prove it without the Dixmier Conjecture.