Let us continue this discussion in chat.

I get that the two are mathematically related by the Wigner map you wrote. What I don't understand is why in the reference I mentionned (and in most textbooks I came across really) they use $A(x,-\frac{i}{\hbar}\partial_x)$ as I wrote in my comment and not $A(x,y) = <x|\hat A|y>$ as you did. This is where my confusion lies. For the specific case of $\hat A=\hat p$ which is written with a Dirac delta function you do end up with a single integral but this doesn't justify the expression for all possible operators. Thanks again for your time and my apologies for the misunderstanding...

Now what still puzzles me regarding the phase-space is what is the difference between $<x|\hat A|y> \equiv A(x,y)$ and $A(x,p)$, or rather, how are the two related? Maybe I'm mistaken and $x$ and $p$ in the Hilbert space description are different from the $x$ and $p$ of the phase-space description.

You're absolutely correct when you say that the momentum is an operator and that I have no trouble understanding. I should have been more precise by saying that the COHEN-TANNOUDJI's book (which is a standard quantum mechanics textbook) writes $<\psi | \hat A | \psi> = \int dx \psi^*(x)A(x,-\frac{i}{\hbar}\partial _x)\psi(x)$ so when I wrote '$p(x)$' I meant $-\frac{i}{\hbar}\partial _x$. But just to be sure, you seem to confirm that in general the mean value of an operator written in terms of the position representation should account for two variables, $x$ and $y$?

Thank you for your answer. If you say that indeed one should write $\hat A = \int dxdy|x><y|<x|\hat A|y>$ then can you would agree that $<\psi|\hat A |\psi> = \int dx dy\psi^*(x)<x|\hat A|y>\psi(y)$ and then coud you explain to me why people usually write it as (for example in the COHEN-TANNOUDJI Quantum mechanics I eq. D28) as $\hat A = \int dx \psi^*(x)A(x,p(x))\psi(x)$ ? This is precisely why I can't make sense of the expression you wrote, what is usually written and the phase-space description which accounts for $x,p$.