The function you defined does not satisfy Kolmogorov's axioms for many reasons. First of all it is not even additive!
A quantum state is, in fact, a generalized probability measure on the orthocomplemented lattice of propositions of the considered quantum system.
In the absence of constraints, like superselection rules or a gauge groups, these propositions or events are represented by the orthogonal projectors in the Hilbert space $H$ of the system.
$$\rho : {\cal L}(H)\to [0,1]$$ The lattice is above indicated by ${\cal L}(H)$.
By definition
(i) $\rho(I)=1$ and
(ii) a suitably $\sigma$-additivity condition holds: (the series of projectors is computed in the strong operator topology)
$$\rho\left( \sum_{j \in \mathbb{N}} P_j \right) = \sum_{j\in \mathbb{N}} \rho(P_j)\quad if \:P_k \in {\cal L}(H)\quad and \quad P_iP_j=0\quad for \quad i\neq j\:.$$
If $H$ is separable with dimension $\neq 2$, then the above probability measures are one-to-one with the statistical operators, i.e., positive, trace-class operators, with unit trace $T: H\to H$.
The afore-mentioned correspondence is that
$\rho$ corresponds to $T$ if and only if $\rho(P)= tr(TP)$.
That is the basic statement of the celebrated Gleason theorem.
Pure states, i.e., unit vectors $\psi$ up to phases, are nothing but the extremal elements $\rho_\psi$ of the convex space of the above probability measures and are of the form $\rho_\psi(P) = \langle \psi| P\psi\rangle$ for every $P \in {\cal L}(H)$.
What is the interplay of these generalized probability measures and the ones in the standard theory by Kolmogorov?
${\cal L}(H)$ includes a plethora of maximal sets ${\cal L}_0$ of pairwise commuting projectors. Each of these maximal sets turns out to have the (abstract) structure of a $\sigma$-algebra when restricting the lattice operations to ${\cal L}_0$. Furthermore, a state $\rho \equiv T_\rho$ becomes a true Kolmogorov measure when restricted to every ${\cal L}_0$.
In that sense every quantum state, i.e., a generalized probability measure,
boils down to a standard probability measure when the space of events is restricted to a "classical" (commuting) space of events.
However:
(1) There are many different maximal sets ${\cal L}_0\subset {\cal L}(H)$ with non-trivial intersection (that is because the commutativity relation is not transitive). They give rise to a very complex structure. In this sense, there are many incompatible "classical" words embedded a quantum system.
(2) In view of an elementary topological obstruction (if $dim H >2$), commonly known as the Bell-Kochen-Speker theorem, there are no sharp probability measures on ${\cal L}(H)$. I mean measures that assume only deterministic values $\rho(P) \in \{0,1\}$ if $P\in {\cal L}(P)$. That is quite different from what happens in classical theories. Quantum theory, from this perspective, encompasses a family of intrinsically undeterministic "classical" theories.
In summary, a quantum state can be equivalently defined as an assignment of a Kolmogorov probability measure $\rho_{k} : {\cal L}_k \to [0,1]$ to every maximal subset of pairwise commuting projectors ${\cal L}_k \subset {\cal L}(H)$, such that the coherence condition is satisfied
$$\rho_k(P) =\rho_{j}(P) \quad if \quad P \in {\cal L}_k\cap {\cal L}_j\:.$$
(See also an old answer of mine Why is the application of probability in Quantum Mechanics fundamentally different from application of probability in other areas?)
