I recently completed an exercise showing that $ H_1(G*H,A) \cong H_1(G,A) \oplus H_1(H,A) $ for $A$ a trivial $G*H$-module, and also proved a similar statement for cohomology. This is exercise 6.2.5 in Weibel's Homological Algebra.
Now I'm going back and seeing if I can prove this in a different way than I originally did. Here is my attempt:
Using the fact that $ H_1(G,A) \cong \frac{G}{[G,G]} \otimes_{\mathbb Z G} A $ for $A$ a trivial $G$-module, we have \begin{align*} H_1(G*H,A) &\cong \frac{G*H}{[G*H,G*H]} \otimes_{\mathbb Z(G*H)} A \\ &\cong \left( \frac{G}{[G,G]} \oplus \frac{H}{[H,H]} \right) \otimes_{\mathbb Z(G*H)} A \\ &\cong \left( \frac{G}{[G,G]} \otimes_{\mathbb Z(G*H)} A \right) \oplus \left( \frac{H}{[H,H]} \otimes_{\mathbb Z(G*H)} A \right) . \end{align*} Does it follow from some kind of change of base theorem that $ \frac{G}{[G,G]} \otimes_{\mathbb Z(G*H)} A \cong \frac{G}{[G,G]} \otimes_{\mathbb Z G} A ? $