It's fairly easy to do this for finite groups. In fact, the functor $R \mapsto R[G]$ is naturally representable by a ring scheme: the underlying set functor is represented by $\mathbb A^n$ where $n = |G|$, and the ring structure comes from the functor of points $R \mapsto R[G]$. Write $Y$ for this ring scheme (say over $\operatorname{Spec} \mathbb Z$).
Now the unit group can be constructed as the closed subset $V \subseteq Y \times Y$ of pairs $(x,y)$ such that $xy = 1$. It is closed because it is the pullback of the diagram
$$\begin{array}{ccc}V & \to & Y \times Y\\\downarrow & & \downarrow \\ 1 & \hookrightarrow & Y\end{array},$$
where the right vertical map is the multiplication morphism on $Y$. This shows that $R \mapsto R[G]^\times$ is representable. It naturally becomes a group scheme, again by the functor of points point of view. $\square$
In the infinite case, this construction doesn't work, because the functor $R \mapsto R[G]$ is not represented by $\mathbb A^G$ (the latter represents the infinite direct product $R \mapsto R^G$, not the direct sum $R \mapsto R^{(G)}$). I have no idea whether the functor $R \mapsto R^{(G)}$ (equivalently, the sheaf $\mathcal O^{(G)}$) is representable, but I think it might not be.
On the other hand, in the example you give of $G = \mathbb Z$, the functor on fields
$$K \mapsto K[x,x^{-1}]^\times = K^\times \times \mathbb Z$$
is representable by $\coprod_{i \in \mathbb Z} \mathbb G_m$, but this does not represent the functor $R \mapsto R[x,x^{-1}]^\times$ on rings for multiple reasons. Indeed, it is no longer true that $R[x,x^{-1}]^\times = R^\times \times \mathbb Z$ if $R$ is non-reduced, nor does $\coprod \mathbb G_m$ represent $R \mapsto R^\times \times \mathbb Z$ if $\operatorname{Spec} R$ is disconnected. These problems do not cancel out, as can already be seen by taking $R = k[\varepsilon]/(\varepsilon^2)$.
