Let $\mathfrak{M}$ and $\mathfrak{J}$ be JBW$^*$-algebras admitting no central summands of type $I_1$ and $I_2,$ and let $Φ: \mathfrak{M} \rightarrow \mathfrak{J}$ be a linear bijection preserving operator commutativity in both directions, that is, $$[x,\mathfrak{M},y] = 0 \Leftrightarrow [Φ(x),\mathfrak{J},Φ(y)] = 0,$$ for all $x,y\in \mathfrak{M}$, where the associator of three elements $a,b,c$ in $\mathfrak{M}$ is defined by $[a,b,c]:=(a\circ b)\circ c - (c\circ b)\circ a$. We prove that under these conditions there exist a unique invertible central element $z_0$ in $\mathfrak{J}$, a unique Jordan isomorphism $J: \mathfrak{M} \rightarrow \mathfrak{J}$, and a unique linear mapping $β$ from $\mathfrak{M}$ to the centre of $\mathfrak{J}$ satisfying $$ Φ(x) = z_0 \circ J(x) + β(x), $$ for all $x\in \mathfrak{M}.$ Furthermore, if $Φ$ is a symmetric mapping (i.e., $Φ(x^*) = Φ(x)^*$ for all $x\in \mathfrak{M}$), the element $z_0$ is self-adjoint, $J$ is a Jordan $^*$-isomorphism, and $β$ is a symmetric mapping too. In case that $\mathfrak{J}$ is a JBW$^*$-algebra admitting no central summands of type $I_1$, we also address the problem of describing the form of all symmetric bilinear mappings $B : \mathfrak{J}\times \mathfrak{J}\to \mathfrak{J}$ whose trace is associating (i.e., $[B(a,a),b,a] = 0,$ for all $a, b \in \mathfrak{J})$ providing a complete solution to it. We also determine the form of all associating linear maps on $\mathfrak{J}$.