Relation between Poisson bracket and commutator. In quantum case, we have commutators. In classical case, we have Poisson bracket. Let $T$ be a Poisson group, $a, b \in \mathbb{C}_q[T].$ It seems that we have 
$$
[a, b]=(q-1)\{a,b\}+o((q-1)^2).
$$
Is this true? When $q\to 1$, we have $[a,b]/(q-1)\to \{a,b\}$.
 A: Your notation entails undefined terms and peculiar notation. I will, instead, use the conventional ones. 
Using the invertible Wigner map from linear operators  Φ  in Hilbert space to functions in phase space f(x,p), one has
$$
f(x,p)= 2  \int_{-\infty}^\infty \text{d}y~e^{-2ipy/\hbar}~ \langle x+y| \Phi [f] |x-y \rangle.
$$
In phase space, one may now compare apples with apples, and indeed, the commutator of two operator maps to the bilinear of the corresponding two functions f and  g,
$$
f \star g - g \star f   = f \, \exp{\left( \frac{i \hbar}{2} \left(\overleftarrow {\partial }_x
\overrightarrow{\partial }_p-\overleftarrow{\partial }_p \overrightarrow {\partial }_x \right) \right)}  \, g -  g \, \exp{\left( \frac{i \hbar}{2} \left(\overleftarrow {\partial }_x
\overrightarrow{\partial }_p-\overleftarrow{\partial }_p \overrightarrow {\partial }_x \right) \right)}  \, f \\
=  2i~~ f  \sin \left ( {\frac{\hbar }{2} (\overset{\leftarrow}{\partial_x}
\overset{\rightarrow}{\partial_p}-\overset{\leftarrow}{\partial_p}\overset{\rightarrow}{\partial_x})} \right ) 
\  g =i\hbar \{f,g\} + O(\hbar^3),
$$
with the classical $\hbar\to 0$  limit you are evoking.
