I know that for vector spaces $V, W$ over a field $K$, we have the following identity : $$ \bigoplus_{k=0}^n \left[ \Lambda^k(V) \otimes_K \Lambda^{n-k}(W) \right] \simeq \Lambda^n(V \oplus W) $$ which holds in a canonical way : for $0 \le k \le n$, the maps $$ \Lambda^k(V) \times \Lambda^{n-k}(W) \to \Lambda^n(V \oplus W), \\ (v_1 \wedge\cdots \wedge v_k,w_{k+1} \wedge \cdots \wedge w_n) \mapsto (v_1 \wedge \cdots \wedge v_k) \wedge (w_{k+1} \wedge \cdots \wedge w_n) $$ factors through the tensor product's universal properties, which gives us a map $\varphi_k : \Lambda^k(V) \otimes_K \Lambda^{n-k}(W) \to \Lambda^n(V \oplus W)$, and since the category of $K$-modules has biproducts, we get a canonical map $$ \varphi = \bigoplus_{k=0}^n \varphi_k : \bigoplus_{k=0}^n \left[ \Lambda^k(V) \otimes_K \Lambda^{n-k}(W) \right] \longrightarrow \Lambda^n(V \oplus W) $$

What if I replace $K$ by a (commutative ring with $1$) ring $A$? Does this map have a kernel? I know it is surjective (it suffices to "look at it", i.e generators of the codomain are obviously all hit by $\varphi$). I don't know if it helps, but I am working in integral domains. In the field case, this is obvious if $V$ and $W$ are finitely generated, since one can use dimension arguments. I wouldn't mind if the above identity held only in the finitely generated case either. If it worked in general it would be cool though.

Note that when trying to compute the inverse map with the universal property, there seems to be a problem... my guess would have been $$ (v_1 + w_1,\cdots,v_n + w_n) \mapsto \bigoplus_{k=0}^n (v_1 \wedge \cdots \wedge v_k) \otimes (w_{k+1} \wedge \cdots \wedge w_n) $$ and this map is obviously $A$-multilinear, but there is a problem with the alternating property ; I get some terms left. I'm guessing this is the wrong map... if there is one.