At first, I attempted to do a proof by showing a homomorphism between the groups, then bijectivity:
First, let us define a function $f∶G \to \{0\}$ by $f(x) = 0$. Note that $f(xy) = 0+0 = f(x)+f(y)$ for all $x, y \in G$. Thus, $f$ is a homomorphism. Suppose for some $x \in G$, $f(x) = e_G$ then $0 = f(x) = e_G$, furthermore $e_G = x+0 = x+eG = x$ thus $f$ is injective. Let $x \in \{0\}$ and $a \in G$. Then set $x = a = 0 \in \{0\}$ thus $f$ is surjective and consequently bijective. Thus, $G \cong \{0\}$.
Then I realized that I was approaching it completely incorrectly (and the proof itself is incorrect). So follows my second attempt after getting the hint that $G$ only has one element $\{e_G\}$:
First, let us define a function $f \colon G \to \{0\}$ by $f(x) = 0$. Note that since $\left| G \right| = 1$ then $G = \{e_G\}$.
Am I overthinking this and I could simply state that since there are the same amount of elements in both groups that they are consequently isomorphic?
*I'm more confused and lost than ever ... Clearly I'm not an expert by any means at Group Theory, so please, be gentle.
In general, from two groups having the same number of elements, one cannot conclude that they are isomorphic. An example for this is $\mathbb{Z}_4$ the cyclic group of order $4$ and $\mathbb{Z}_2 \oplus \mathbb{Z}_2$ the Klein-4-group.
However, in your case, since we know that every group has at least one element (the identity) i.e. the trivial group is a subgroup of $G$ and we know that $G$ has only one element, we can conclude that $\{e\}$ and $G$ coincide. Here $e$ denotes the identity element of a group.
If you want to have an explicit isomorphism between $G$ and $\{e\}$ then you can do this by $x \mapsto e$ for every $x \in G$.