Wigner transform, convolution, and poles Let
\begin{equation}
\int\mathrm{d}z~ A(x,z) B(z,y) = \delta(x - y).
\end{equation}
Taking Wigner transform of both sides we readily obtain
\begin{equation}
A^W(X,p) \star B^W(X,p) = 1,
\end{equation}
where $\star$ is the Weyl–Groenewold (or Moyal) product. Now, to leading order in the gradient expansion (or in $\hbar$ if you prefer), $$A^W(X,p) \star B^W(X,p) = A^W(X,p) B^W(X,p) + O(\hbar),$$ which suggests that poles of $A^W$ coincide with zeros of $B^W$ and vice versa.
My question is: can this statement be generalized to the full case, i.e., without approximating the $\star$-product? My guess would be the answer is 'no' and that there is a simple counterexample showing this already on a Poisson-bracket level. But perhaps I am wrong?
 A: I don't quite know, but I'd agree with your bet that the general answer is "no". I'll just jot down a few remarks and an unsatisfactory example that is easier to compute with; however, it is based on the "crypto-semiclassical" oscillator that most seasoned professionals appreciate is quite unrepresentative of generic QM situations, despite its clarity and usefulness.

*

*The star product is just a (von Neumann-Baker) convolution in phase space,
$$A^W(x,p)\star B^W(x,p)\\ = {1\over \hbar ^2 \pi^2}\int dp^{\prime}  
dp^{\prime\prime}  dx' dx''  ~A^W(x',p')~B^W(x'',p'')   \\
 \times    \exp \left({-2i\over \hbar} 
\left( p(x'-x'') + p'(x''-x)+p''(x-x') \right )\right) , $$
not qualitatively different than your Hilbert-space integral kernel convolution gambit expression. These convolutions, in sharp contrast to local multiplication composition of functions, smear them and unzip them and rezip them nonlocally, so their extended structure and their variations around a point add and cancel in very different ways, incur local imbalances "loans" of their value to be repaid elsewhere in their domain; so little can be said about the structure around their zeros and singularities, except by heavy-duty professional functional analysts, which I am not.  Convolutions probe  variation  of functions, and not just local features at points.


*The ℏ is not quite an irrelevant aside: By virtue of dimensionality, it amounts to the scale of relevant variation of the (Weyl-symbol) functions involved: $\sqrt \hbar$ must divide x and p, and therefore multiply their corresponding gradients. These variations are where all the "action" is in QM. For example, the Wigner functions of stationary pure states of the oscillator depend ferociously nonanalytically on it. I really don't want to send you off on a tangent chasing irrelevancies by pointing you to the "wall-of-shame" papers that misconstrue this fact. (However, lest you went on an unjustified  wild-goose chase  in Takahashi's paper, his section 2 misconstrues the classical limit.)


*As a lark, you might consider  a (poor, unrepresentative) example based on the rescaled and nondimensionalized oscillator, $H^W=p^2+ x^2$. In Hilbert space,
$$
\hat A = 1\!\!1 -\hat H, \qquad \hat 
B= \sum_{k=0}^\infty \hat H ^k=  1\!\!1 +\hat H + \hat H^2+\hat H^3+ \ldots \\ \leadsto \qquad \hat A \hat B=  1\!\!1,
$$
so
$$
A(x,z)= \langle x|1\!\! 1-(\hat p^2+ \hat x^2)  |z\rangle \\ = \delta (x-z) -x^2\delta (x-z) + \hbar^2 \delta '' (x-z),
$$
etc. Before indulging in star products, see how these Hilbert-space nonlocal expressions address your question satisfactorily.
In phase space, this particular expression $A^W\star B^W$ is easy to see to be real and analytic in ℏ, a very unrepresentative circumstance. Given the Bopp shift
$$
H^W \star H^W= \left ( (x+i\hbar{\partial_p\over 2})^2+(p-i\hbar{\partial_x\over 2})^2 \right )(x^2+p^2)\\ 
=(x^2+p^2) (x^2+p^2-\hbar^2),  
$$
etc, you see
$$1=  (1-H^W)\star \bigl ( 1+H^W+H^W\star H^W ~+H^W\star H^W\star H^W +\ldots \bigr ) \\ =\left (1-H^W +\frac{\hbar^2}{4}(\partial_x^2+\partial_p^2 )\right ) ~ \bigl ( 1+H^W+H^W\star H^W ~+H^W\star H^W\star H^W +  \ldots \bigr ) 
$$
dictates that all $O(\hbar)$ terms cancel among themselves in the equation, so they cancel their HW (the effective variable$^\natural$) singularities and zeros order by order in ℏ. Note the recursion relation that follows from it!  But, as indicated, this (analyticity in ℏ) is a scandalously smooth and untypical circumstance. A real  condign example should transcend this...

$^\natural$ So, defining $z\equiv H^W$, you just have
$$
(1-z+\hbar^2 \partial_z+ \hbar^2 z\partial_z^2)   B^W(z, \hbar)=1,
$$
if you like ODEs, whose  singularities you may study better. So
$$
(A^W+ \hbar^2 \partial_z ~ z~\partial_z)(a_0+ \sum_{n=1}^\infty \hbar^{2n} a_n )=1~~~~\leadsto\\
a_n= -\frac{1}{1-z} \partial_z ~ z~\partial_z ~a_{n-1} , ~~~a_0=\frac{1}{ A^W}=\frac{1}{1-z}.  
$$
Thus $B^W= {1\over 1-z}-\hbar^2 {1+z\over (1-z)^4}+O(\hbar^4)$, etc. Here, the poles are unshifted, but note the orders: The ℏ terms cancel among themselves, not with the classical part.
