$\mathbb{Z}G$ is necessarily a subgroup of $G$, or am I missing something? Let $\mathbb{Z}G=\left\{\sum\limits_{g\in G}n_gg\mid n_g\in\mathbb{Z},g\in G\right\}$, where if $G$ is infinite, we only consider finite formal sums of elements of $G$ with coefficients in $\mathbb{Z}$.
I feel like $\mathbb{Z}G$ is necessarily a subgroup of $G$ by the closure of $G$. 
Is that right? Or am I missing something?
If I am wrong, can you think of a counterexample?
 A: ${\mathbb Z}G$ is not even a subset of $G$, let alone a subgroup. 
Edit: It might be instructive to compute ${\mathbb Z}G$ in the case when $G=\{e\}$, the trivial group. Then ${\mathbb Z}G$ naturally identified with the set 
${\mathbb Z}$ of integers. Clearly, ${\mathbb Z}G$ cannot be even embedded in $G$ as a subset. What you do have are the natural maps (for general $G$)
$$
\epsilon: {\mathbb Z}G \to \mathbb Z
$$ 
(called the augmentation homomorphism of $\mathbb Z$-modules) 
$$
\epsilon: \sum_{g\in G} n_g g \to \sum_{g\in G} n_g\in \mathbb Z
$$
and the inclusion map
$$
\iota: G\to {\mathbb Z}G
$$
sending each $g\in G$ to the element $1 g$ of the group ring (where $1$ is in ${\mathbb Z}$). 
A: You are misunderstanding what $n_gg$ means. When doing group rings, it's customary to use the multiplicative notation for the operation on the group.
The “correct” definition of elements in a group ring is like

$\mathbb{Z}G$ is the set of all functions $\alpha\colon G\to \mathbb{Z}$ with finite support, that is, such that $\{g\in G:\alpha(g)\ne0\}$ is finite.

The addition in the group ring is defined componentwise, while multiplication is defined by
$$
\alpha\beta\colon g\mapsto \sum_{xy=g}\alpha(x)\beta(y)
$$
If $\varphi_g$ is the function defined by
$$
\varphi_g(x)=\begin{cases}1 & \text{if $x=g$}\\0 & \text{if $x\ne g$}\end{cases}
$$
then we can identify $g$ with $\varphi_g$ and so justify the notation
$$
\sum_{g\in G}n_gg
$$
because, for $\alpha\in\mathbb{Z}G$ we have
$$
\alpha=\sum_{g\in G}\alpha(g)\varphi_g
$$
So you see that $\mathbb{Z}G$ is not a subset of $G$.
