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), 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, you need two obvious properties:
$D\in\mathcal{D}^n_A(B) $ if and only if $\mathrm{S}^{n+1}(D)=0$;
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 $\mathcal{D}^n_A(M)\equiv \mathcal{D}^n_A(B)\otimes_B\mathrm{End}_B(M)$.
If the modules are of a "bad" kind, then I wouldn't know how to deal with them!