Algebraically, the relation $[a,b] = iħ\{a,b\}$ isn't just a "correspondence" or something that holds "only in the limit", but is true as is. More precisely, when taken with the definition $a·b ≡ ½(ab + ba)$, the result is a decomposition
$$ab = a·b + \frac{iħ}{2} \{a,b\},$$
and an algebra satisfying the following properties:
$$
a·b = b·a, \hspace 1em \{a,b\} = -\{b,a\},\\
a·(b·c) - (a·b)·c = \left(\frac{ħ}{2}\right)^2 \{b,\{a,c\}\},\\
\{a,b·c\} = \{a,b\}·c + b·\{a,c\},\\
\{a,\{b,c\}\} = \{\{a,b\},c\} + \{b,\{a,c\}\}.
$$
Apart from the defect in the associativity property by the small factor $-(½ħ)^2$, this is just the same kind of algebra as for a classical Poisson manifold; i.e. a Poisson algebra. So, the operator algebra is a deformation of a Poisson algebra. You could just as well write everything algebraically in this way - without the use of complex numbers at all. The rendering of it as an associative, but now non-commutative, complex algebra is just a way of packing together the structure of a real commutative, but non-associative product with a Lie bracket. The only place where the quantum nature is set apart from the classical nature is in the identity for the associativity defect.
This is connected to the Weyl "operator ordering" ($W[⋯]$ = the average taken over all permutation of the basic operators in "⋯") as follows:
$$
W[ab] ≡ \frac{ab + ba}{2} = \frac{a·b + b·a}{2},\\
W[abc] ≡ \frac{abc + acb + bac + bca + cab + cba}{6} = \frac{a·(b·c) + b·(c·a) + c·(a·b)}{3},$$
and the cubic combinations of basic operators $a$, $b$, $c$, may be expressed as
$$abc = \frac{(a·b)·c + a·(b·c)}{2} + iħ \frac{a·\{b,c\} - b·\{c,a\} + c·\{a,b\}}{2} - ħ^2 \frac{\{\{a,b\},c\} + \{a,\{b,c\}\}}{8}.$$
A limited form of associativity, for powers of the same factor, holds:
$$\left(a·b^m\right)·b^n - \left(a·b^n\right)·b^m = \frac{ħ^2}{4} \{\{b^m,b^n\},a\} = \frac{h^2}{4} \{0,a\} = 0 \hspace 1em (m, n = 0, 1, 2, ⋯).$$
It also bears to note that the Lie bracket also satisfies the product rule for the complex product, itself:
$$\begin{align}
\{a,bc\} &= \{a,b·c\} + \frac{ih}{2} \{a,\{b,c\}\}\\
&= \{a,b\}·c + b·\{a,c\} + \frac{ih}{2} (\{\{a,b\},c\} + \{b,\{a,c\}\})\\
&= \left(\{a,b\}·c + \frac{ih}{2} \{\{a,b\},c\}\right) + \left(b·\{a,c\} + \frac{ih}{2} \{b,\{a,c\}\}\right)\\
&= \{a,b\}c + b\{a,c\}.
\end{align}$$
Jordan-Lie and Jordan-Banach Algebras:
https://planetmath.org/JordanBanachAndJordanLieAlgebras
Another sense in which the correspondence to the Poisson bracket holds true is that the underlying Hilbert space is, itself, already a Poisson manifold, with the imaginary part of the inner product connected to the Poisson bracket. But I'm still not entirely clear, yet, on what the exact relation is between its Poisson bracket and the above-mentioned Lie bracket.
I'll have to work through this, here.
More precisely, quantum states are not actually non-zero Hilbert space vectors. Instead, a vector $v ∈ ℌ - \{0\}$ in a Hilbert space $ℌ$ gives rise to a pure state that may be identified uniquely as $W_v = |v〉〈v|/|v|^2$, where $|v| = \sqrt{〈v,v〉}$. This is a projective geometry since $W_v = W_{-v}$ and, in fact, $W_v = W_{λv}$ for any real or complex $λ ≠ 0$. Mixed states are convex unit norm sums or integrals of pure states.
The inner product of two states is
$$〈W_v, W_{v'}〉 = \frac{〈v,v'〉}{|v||v'|},$$
which satisfies the identity
$$|〈W_v, W_{v'}〉|^2 = Tr(W_v W_{v'}),$$
and can, thus, be considered as a complex square root of the states' overlap.
This extends, bi-linearly to an inner product $〈W, W'〉$ the full space of pure and mixed states, $W$, $W'$.
The Poisson bracket arises, via a Poisson tensor $Ω(W, W')$, as the complex part of the inner product
$$〈W, W'〉 = G(W, W') + i Ω(W, W') = 2ħ〈W, W'〉,$$
the real part giving you a metric $G(W, W')$ for a Kähler manifold, with the constraint $G(W, W') = 2ħ$ giving rise to the re-phase symmetry of the Hilbert space $ℌ$, itself, as a gauge symmetry.
The expectation value of an operator $a$ in state $W_v$ is $\widetilde{a}(W_v) = Tr(W_v a) = 〈v|a|v〉/〈v,v〉$, so operators are actually functions over this space, and their Poisson bracket is directly connected to the Lie Bracket as
$$\{\widetilde{a}, \widetilde{b}\} = \widetilde{\{a,b\}}.$$
That's close to how N. P. Landsman treats it in his 1998
Mathematical Topics Between Classical and Quantum Mechanics
https://link.springer.com/book/10.1007/978-1-4612-1680-3
in his section 2.5, except his treatment was more cumbersome, because he failed to make the direct identification of $W_v$ as the state for the Hilbert space vector $v$. Ashtekar and Shilding also discussed the underlying geometry in:
Geometric Formulation Of Quantum Mechanics
https://arxiv.org/pdf/gr-qc/9706069v1.pdf
(1997 June 21)
So, the Lie bracket $\{a,b\}$ actually is a Poisson bracket, when treated as a state space function; not merely something that has a Poisson bracket as a classical limit or corresponds to a Poisson bracket. Instead, what you find, here, is that the classical Poisson bracket, is the classical limit of the quantum Poisson bracket:
$$\lim_{ħ → 0}\widetilde{\{a,b\}} = \lim_{ħ → 0}\{\widetilde{a},\widetilde{b}\} = \left\{\lim_{ħ → 0}\widetilde{a},\lim_{ħ → 0}\widetilde{b}\right\},$$
and the entire infrastructure of the algebras, manifolds and brackets are passing over into the limit (or in the reverse direction: arising as deformations of their respective classical limits), not just individual objects.