In a 2D CFT, the commutator of two operators
$$A_i=\oint a_i(z)dz$$
can be given by $$[A_1,A_2]=\oint_0dw\oint_wdza_1(z)a_2(w)$$
where the $z$ integral is taken over a contour around $w$ and the $w$ integral is taken over a contour around the origin, and $a_i(z)$ are holomorphic operators.
What does this look like in the classical limit? When the commutator is replaced with the Poisson bracket $$[\cdot,\cdot]\to i\hbar\{\cdot,\cdot\}.$$ What happens to the integral on the right hand side?
If $$\hat{a}(z)~=~\sum_n z^{-n-h_a}\hat{a}_n
\quad\text{and}\quad
\hat{b}(w)~=~\sum_m w^{-m-h_b}\hat{b}_m,\tag{1}$$
or conversely,
$$ \hat{a}_n~=~\oint_0 \frac{\mathrm{d}z}{2\pi i}z^{n+h_a-1}\hat{a}(z)
\quad\text{and}\quad
\hat{b}_m~=~\oint_0 \frac{\mathrm{d}w}{2\pi i}w^{m+h_b-1}\hat{b}(w),\tag{2}
$$
then OP is essentially interested in the classical limit $\hbar\to 0$ of the OPE formula
$$ [\hat{a}_n,\hat{b}_m]~=~\oint_0 \frac{\mathrm{d}w}{2\pi i}\oint_w \frac{\mathrm{d}z}{2\pi i}z^{n+h_a-1} w^{m+h_b-1}{\cal R}\hat{a}(z)\hat{b}(w).\tag{3} $$
Here the symbol ${\cal R}$ denotes radial ordering,
$${\cal R} \hat{a}(z)\hat{b}(w)~:=~\left\{ \begin{array}{rcl} \hat{a}(z)\circ\hat{b}(w)&{\rm for}&|z|>|w|, \cr \hat{b}(w)\circ\hat{a}(z)&{\rm for}&|w|>|z|.\end{array}\right. \tag{4}$$
The symbol ${\cal R}$ itself is often implicitly implied in CFT texts. Concerning the proof of the formula (3), see e.g. this Phys.SE post.
It seems OP is already well aware of the classical limit $\hbar\to 0$ of the commutator on the LHS of eq. (3) in terms of the Poisson bracket, cf. e.g. this Phys.SE post. Obviously the classical limit $\hbar\to 0$ of the RHS of eq. (3) has to yield precisely the same result.
One idea for further insides is to replace the operators $\hat{a}(z)$, $\hat{b}(w)$ and composition $\circ$ with symbols/functions $a(z)$, $b(w)$ and the star product $\star$, and then expand in $\hbar$.
