Not so immediately, but rather easily (as far as the modules are of a "good" kind, i.e., projective and finitely generated). It all boils down to prove that, if
$$
D\in\mathcal{D}^n_A(B):=\mathcal{D}^n_A(B,B)\, ,
$$
then the map
$$
B\times\cdots\times B\ni(b_1,\ldots,b_n)\stackrel{\mathrm{S}^n(D)}{\longrightarrow}[b_1\cdots [b_n,D]\cdots]\in\mathrm{End}_B(B)\equiv B
$$
is $A$-multilinear and, with respect to each entry, it fulfils the Leibniz rule. Since a derivation factors through a $B$-linear map on the $B$-module of differential one-forms $\Lambda^1B$ (in analogy with the "jet modules" I mentioned [before][1]), the "$n^\textrm{th}$ symbol" $\mathrm{S}^n(D)$ of $D$ may be also thought of as an element of the $B$-module
$$
\mathrm{Sym}^n(\Lambda^1B) \, .
$$
Now, to check your claim (EDIT: which is certainly true for $M=B$, see Michael Bachtold's comments below), you need two obvious properties:

1) $D\in\mathcal{D}^n_A(B) $ if and only if $\mathrm{S}^{n+1}(D)=0$;

2) if $D_1\in\mathcal{D}^{n_1}_A(B) $ and $D_2\in\mathcal{D}^{n_2}_A(B) $, then $\mathrm{S}^{n_1+n_2}(D_1\circ D_2)=\mathrm{S}^{n_1}(D_1)\cdot \mathrm{S}^{n_2}(D_2)$, where   "$\cdot$" is the product in the graded commutative algebra $\mathrm{Sym}(\Lambda^1B)$.

The general case, i.e., when $D\in \mathcal{D}^n_A(M):=\mathcal{D}^n_A(M,M)$, can be reduced to the previous one by using the identifications
$$
\mathcal{D}^n_A(M)\ni D\longleftrightarrow \widehat{D}\in\mathrm{Hom}_B(M, \mathcal{D}^n_A(B,M))\, ,
$$
where $\widehat{D}(m)(b):=D(bm)$, and $\mathcal{D}^n_A(B,M)\equiv \mathcal{D}^n_A(B)\otimes_B M$, which imply  $\mathcal{D}^n_A(M)\equiv \mathcal{D}^n_A(B)\otimes_B\mathrm{End}_B(M)$, but commutativity is not guaranteed (again, see comments below).

If the modules are of a "bad" kind, then I wouldn't know how to deal with them!



  [1]: https://mathoverflow.net/a/210975/22606