I do not know any reference for your specific problem. However, most books or lecture notes on set theory, especially, in the chapters that discuss cardinality should have sufficient information for you to deduce what you need. You would at least need to know the Schröder–Bernstein Theorem to prove your statement. Below is my proof of your statement.
Presumably, for each $f:E\to\mathbb{Z}$, the support $\text{supp}(f)$ of $f$ is the set of elements $x$ of $E$ such that $f(x)\neq 0$. For each function $f:E\to \mathbb{Z}$, identify $f$ with the reduction $\hat{f}:\text{supp}(f)\to \mathbb{Z}_{\ne 0}$, where $\hat{f}(x):=f(x)$ for every $x\in\text{supp}(f)$. From this identification, we have
$$G=\bigcup_{F\in\mathcal{P}_\text{fin}(E)}\,\text{Maps}(F,\mathbb{Z}_{\neq 0})\,,$$
where $\mathcal{P}_\text{fin}(S)$ is the set of all finite subsets of a set $S$, and $\text{Maps}(A,B)$ denotes the set of all functions from a set $A$ to a set $B$. It is known that
$$\big|\mathcal{P}_\text{fin}(S)\big|=|S|$$
for every infinite set $S$. Furthermore, if $A$ is an nonempty finite set and $B$ is an infinite set, then
$$\big|\text{Maps}(A,B)\big|=|B|^{|A|}=|B|\,.$$
Thus,
$$G=\{\emptyset\}\cup\bigcup_{\substack{F\in\mathcal{P}_\text{fin}(E)\\ |F|>0}}\,\text{Maps}(F,\mathbb{Z}_{\neq 0})$$
with
$$\left|\bigcup_{\substack{F\in\mathcal{P}_\text{fin}(E)\\ |F|>0}}\,\text{Maps}(F,\mathbb{Z}_{\neq 0})\right|=\left|\mathcal{P}_{\text{fin}}(E)\times \mathbb{Z}_{\neq 0}\right|=|E|\,|\mathbb{Z}|=|E|\,.$$
That is, $|G|=1+|E|=|E|$. Therefore, there exists a bijection between $E$ and $G$.
Remark. Let $P$ be a pointed set, say, $0\in P$ is the point. For each $f:E\to P$, we define $\text{supp}(f)$ to be, as before, the set of $x\in P$ such that $f(x)\neq 0$. Similarly, if $P$ is an infinite set and $|E|\geq |P|$, then we have $$|E|=\Big|\big\{f:E\to P\,\big|\,\text{supp}(f)\text{ is finite}\big\}\Big|\,.$$
Below is an answer by HellaSurvivor in a deleted thread. If the current thread is opened, I would like to invite HellaSurvivor to copy the answer here. I am putting HellaSurvivor's here to try to preserve the answer. (There are other answers both here and in the deleted thread, but since the authors deleted their own answers, I am not going to copy their answers here.)
HellaSurvivor's Answer. I don't know of a reference for your exact claim, but the theorem is implicit from theorems in the first chapter of Kunen's "Set Theory". The following proof requires choice, and I suspect the theorem is false without it.
Obviously $|E| \leq |G|$. Send each $e \in E$ to its characteristic function to witness the injection.
Conversely, write $G_n$ for those elements of $G$ with support of size exactly $n$, so $G$ is the (disjoint) union of the $G_n$. Then each $G_n$ can be identified with a subset of $E^n \times \mathbb{Z}^n$ by coding each $f \in G_n$ by the $n$ elements in the support (in some arbitrary order) plus their values. So $$|G_n| \leq |E^n| \cdot |\mathbb{Z}^n| = |E|\,,$$ since $E$ is infinite.
Thus,
$$|G| = \left| \bigcup_\omega G_n \right| = \sum_\omega |G_n| \leq \sum_\omega |E| = \aleph_0 \cdot |E| = |E|\,.$$
