What's wrong in this "counterexample" of the Moyal´s product associativity?

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I have been working in the phase space formulation of quantum mechanics, and stumbled upon the following problem: Consider the phase space functions :

$e^{ip\epsilon /\hbar}$, $e^{-ip\epsilon /\hbar}$ and $W(x,p)$,

where $e^{\pm ip\epsilon /\hbar}$ are asociated with forward and backwards quantum tranlsation operators, $W(x,p)$ is a Wigner pseudoprobability function, and as usual $x$ stands for position and $p$ for momentum. I need to compute the quantity

$e^{ip\epsilon /\hbar} * W(x,p)* e^{-ip\epsilon /\hbar}$

Where * represents the Moyal product,

$A(x,p)*B(x,p)= A(x,p) e^{i \Lambda \hbar/2} B(x,p)$

$\Lambda= \overleftarrow{\partial_p}$ $\vec{\partial_x} $ - $\overleftarrow{\partial_x}$ $\vec{\partial_p} $

On one hand, using Moyal´s product associativity and computing

$e^{ip\epsilon /\hbar} * W(x,p)* e^{-ip\epsilon /\hbar}=e^{ip\epsilon /\hbar} * [W(x,p)* e^{-ip\epsilon /\hbar}]$

$=e^{ip\epsilon /\hbar}*[e^{-ip\epsilon /\hbar}e^{-\epsilon\partial_x/2}W(x,p)]$

$=e^{-\epsilon\partial_x}W(x,p)$

Which I verified using other methods its the correct result. On the other hand, using the following property of Moyal´s product

$A(x,p) e^{\Lambda \hbar/2} B(x,p)= B(x,p) e^{-\Lambda \hbar/2} A(x,p)$

I can rearrange:

$e^{ip\epsilon /\hbar} * W(x,p)* e^{-ip\epsilon /\hbar}=e^{ip\epsilon /\hbar} e^{\Lambda \hbar/2} e^{-ip\epsilon /\hbar}e^{-\Lambda \hbar/2} W(x,p)$

Now invoking asociativity,

$=[e^{ip\epsilon /\hbar} * e^{-ip\epsilon /\hbar}]e^{-\Lambda \hbar/2} W(x,p)$ $=[1] e^{-\Lambda \hbar/2} W(x,p)$ $=W(x,p)$

Clearly there must be a failure in the above reasoning, or I have misused the product´s properties somewhere, but I cannot find the mistake. Can anyone point me out where I screwed up?