This is probably not really an answer to this question, but there are two different context I know where deformation quantization produces something not exactly associative, but associative in a larger sense. This doesn't encompass all cases covered by the question, but might give some hints.

a) First, there are so-called quasi-Poisson $G$-manifolds for which the bivector doesn't satisfy the Jacobi identity. Here is the setup. A quasi-Poisson $G$-manifold ($G$ being a Lie group) is a $G$-manifold $M$ together with a bivector $\pi$ and an element $Z\in\wedge^3(\mathfrak{g})^G$ ($\mathfrak{g}$ being the Lie algebra of $G$) and such that $[\pi,\pi]=\vec{Z}$ (here $x\mapsto\vec{x}$ send an element of $\mathfrak{g}$ to the associated fundamental vector field of the action). In other words, the default of the Jacobi identity is determined by $Z$.

In this context one can chose an associator $\Phi=1+\hbar Z+\cdots\in(U(\mathfrak{g})^{\otimes3})^G[[\hbar]]$, and consider the monoidal category $\mathcal C_\hbar$ of $\mathfrak{g}[[\hbar]]$-modules whith non-trivial natural associativity isomorphism being given by $\Phi$. A quantization of $\pi$ is then a product on $C^\infty(M)[[\hbar]]$ deforming the usual product of fonctions **as an associative monoid in $\mathcal C_\hbar$**.

Since the forgetful functor $\mathcal C_\hbar\to Vect_\hbar$ is not monoidal, then the quantization doesn't seem associative. But secretely, it is.

b) The second situation I have in mind in which one gets non-associative deformation is probably more in the spirit of the what is alluded to in the question. It is the one of twisted Poisson manifolds, where the default of satisfying the Jacobi identity is rather controlled by a closed $3$-form $H$ on $M$ (sometimes refered to as a *magnetic charge* in more physics papers). Here is what happends:

locally, this $3$-form is exact, and thus the twisted Poisson structure is isomorphic to a genuine non-twisted one.

on intersections, one can prove that we do have compatibilities... which unfortunately don't glue well (i.e. they don't satisfy the expected cocycle identity).

it turns out that things do glue to higher order (getting a gerby version of a sheaf of Poisson algebras). The glueing data is determined by a Cech $3$-cocycle, which represents the same cohomology class in $H^3(M,\mathbb{R})$ as $H$.

These gadgets can be naturally deformed to so-called *algebroid stacks*, which are a generalization of sheaves of algebras, satisfying a higher cocycle condition (sometimes called tetrahedron equation). They can also be seen as linear versions of Giraud's gerbes. A lot of people worked on that in the algebaic and holomorphic contexts: Kontsevich, Yekutieli, Van den Bergh, Halbout and myself, Bressler--Gorokhovsky--Nest--Tsygan, Kashiwara--Schapira, etc... In these contexts these gadgets are necessary for proving existence theorems for deformation quantization of algebraic/holomorphic Poisson structures (indeed, glueing property are known to be harder to handle in these contexts than in the differentiable case).

This approach is less known in the differentiable context, though there are references (I guess it would be fruitful to revisit this paper in the light of all the huge work that has been done in the algebraic and holomorphic cases).

I'm probably biaised, but it seems to me that all the non-associative structures appearing in deformation quantization are secretely associative in a way: either associativity is satisfied up to homotopy (like in the case of deformation quantization with branes, after Cattaneo--Felder), or it is satisfied in an appropriate monoidal category (like for quasi-Poisson manifolds), or it is satisfied locally but glue in a non-trivial way (like in the twisted Poisson situation), etc...

I don't pretend at all that this is all understood. But I would be tempted to say that a good strategy could be to search for hidden associativity whenever one encounters non-associative products of observables in deformation quantization problems.

3more comments