From nothing to a group

First of all, if $E$ is empty, then $E$ cannot be endowed with a group structure for lack of a neutral element.
Consider $E \neq \varnothing$ and let $n \in \mathbb{N}^{*}$ denote the cardinality of $E$. Since the group $(\mathbb{Z}/n\mathbb{Z},+)$ has the same cardinality as $E$, there exists a bijection from $E$ to $\mathbb{Z}/n\mathbb{Z}$, which we shall denote by $\varphi$.
We then consider the map $\tau : E \times E \longrightarrow E$ defined as follows:
$$\forall(x,y) \in E^2, \quad \tau(x,y) = \varphi^{-1} \left( \varphi(x) + \varphi(y)\right) \;.$$We shall then verify that $(E,\tau)$ is a group.
Let $\overline{0}$ denote the neutral element of $(\Z/n\Z,+)$, and set
$$e:=\varphi^{-1}(\overline{0})\in E.$$

Let us first prove associativity. Let $(x,y,z)\in E^3$. By definition of $\tau$, we have
$$\varphi(\tau(x,y))=\varphi(x)+\varphi(y).$$ Therefore
$$\begin{aligned} \varphi\bigl(\tau(\tau(x,y),z)\bigr) &=\varphi(\tau(x,y))+\varphi(z)\\ &=(\varphi(x)+\varphi(y))+\varphi(z)\\ &=\varphi(x)+(\varphi(y)+\varphi(z))\\ &=\varphi(x)+\varphi(\tau(y,z))\\ &=\varphi\bigl(\tau(x,\tau(y,z))\bigr). \end{aligned}$$ As $\varphi$ is injective, it follows that
$$\tau(\tau(x,y),z)=\tau(x,\tau(y,z)).$$ Thus $\tau$ is associative.

Let us verify the existence of a neutral element. For every $x\in E$,
$$\varphi(\tau(e,x)) = \varphi(e)+\varphi(x) = \overline{0}+\varphi(x) = \varphi(x),$$ hence, by injectivity of $\varphi$,
$$\tau(e,x)=x.$$ Similarly,
$$\varphi(\tau(x,e)) = \varphi(x)+\varphi(e) = \varphi(x)+\overline{0} = \varphi(x),$$ whence
$$\tau(x,e)=x.$$ Thus $e$ is a two-sided neutral element for $\tau$.

Finally, let us verify the existence of an inverse for every element. Let $x\in E$. In the group $(\Z/n\Z,+)$, the element $\varphi(x)$ admits the additive inverse $-\varphi(x)$. Set
$$y:=\varphi^{-1}(-\varphi(x))\in E.$$ Then
$$\varphi(\tau(x,y)) = \varphi(x)+\varphi(y) = \varphi(x)-\varphi(x) = \overline{0} = \varphi(e),$$ so, by injectivity of $\varphi$,
$$\tau(x,y)=e.$$ Similarly,
$$\varphi(\tau(y,x)) = \varphi(y)+\varphi(x) = -\varphi(x)+\varphi(x) = \overline{0} = \varphi(e),$$ whence
$$\tau(y,x)=e.$$ Thus every element of $E$ admits a two-sided inverse for $\tau$.

The three axioms are therefore satisfied; consequently, $(E,\tau)$ is a group (and it is even abelian).