Let G be a group, and $P_n=\mathbb{Z[G^{n+1}]}$ be the group ring of $G^{n+1}$. We can turn $P_n$ into a $G$-module by letting multiplication by $g \in G$ take $(g_0,\dots,g_n)$ to $(gg_0,\dots,gg_n)$.
Apparently, it's not hard to see that $P_n = \mathbb{Z[G]} \otimes_\mathbb{Z} Q_n$, where $Q_n$ is a free abelian group generated by $(e,b_1,b_2,\dots,b_n)$ over all $b=(b_1,b_2,\dots,b_n) \in G^n$.
What's the isomorphism? It seems a bit arbitrary to bring up the tensor product out of nowhere.
(It also seems strange to define $Q_n$ using generators $(e,b_1,b_2,\dots,b_n)$ instead of just $(b_1,b_2,\dots,b_n)$, but that's essentially what they did albeit with more complicated notation, where I read this from: http://wwwf.imperial.ac.uk/~anskor/Algebra%20IV/algebraIV.pdf, page 32, Lemma 4.7.)
Remark: It looks like they have an error in their proof, namely in the second formula they give for $P_n(a)$. Surely a cannot be indexed over on the RHS, and still appear on the LHS. I'm too baffled to even tell if they mean $P_n = \mathbb{Z[G]} \otimes_\mathbb{Z} Q_n$ as an isomorphism of abelian groups, or of $G$-modules. Which is it?
Let $G$ and $H$ be groups. Then there is a ring isomorphism $$\newcommand{\Z}{\Bbb Z}\phi:\Z[G]\otimes_\Z\Z[H]\to\Z[G\times H]$$ defined by $$\phi(g\otimes h)=(g,h)$$ where $g\in G$, $h\in H$ and $(g,h)$ is an element of $G\times H$.
Your isomorphism is the case $H=G^n$, your $Q_n$ being $\Z[G^n]$.