The first-order commutator of an associative deformation is Poisson

For any associative product, the commutator $[a,b{]}_{\star }=a\star b-b\star a$ satisfies the Jacobi identity. Expanding $[[f,g{]}_{\star },k{]}_{\star }$ gives $(f\star g)\star k-(g\star f)\star k-k\star (f\star g)+k\star (g\star f)$. After reassociating each term, this is $f\star g\star k-g\star f\star k-k\star f\star g+k\star g\star f$; the two cyclic rotations are $g\star k\star f-k\star g\star f-f\star g\star k+f\star k\star g$ and $k\star f\star g-f\star k\star g-g\star k\star f+g\star f\star k$. Their sum is zero by pairwise cancellation.

Associativity also gives the exact derivation identity $[f,g\star k{]}_{\star }=[f,g{]}_{\star }\star k+g\star [f,k{]}_{\star }$: expanding both sides and reassociating the four triple products leaves the same expression. Comparing coefficients of $\mathrm{\hslash }$ gives $\beta (f,gk)=\beta (f,g)k+g\beta (f,k)$.

Since $fg=gf$, the coefficient of $\mathrm{\hslash }$ in $[f,g{]}_{\star }$ is exactly $\beta (f,g)$. The coefficient of ${\mathrm{\hslash }}^2$ in the commutator Jacobi identity from step 1.1 is $\beta (\beta (f,g),k)+\beta (\beta (g,k),f)+\beta (\beta (k,f),g)=0$. Skew-symmetry of $\beta$ converts this to the Jacobi identity in [F2].

By definition, $\beta$ is real-bilinear and skew-symmetric. The derivation rule in step 1.2 and skew-symmetry give the derivation rule in the first entry; step 2.1 gives Jacobi. These are exactly the Poisson-algebra axioms in [F2].