Unless I was mistaken somewhere in the proof, the following claim should hold:
Claim. Let $G, \tilde G$ be groups. Then, $G\cong\tilde G$ if and only if there are a set $S$ and two bijections, $f\colon S\to G$ and $\tilde f\colon S\to \tilde G$, such that:
$$\tilde f^{-1}(\tilde f(s)\tilde f(t))=f^{-1}(f(s)f(t)), \space\space\forall s,t \in S\tag 1$$
(Operation symbol is omitted in both groups.) As a first test of $(1)$, I'd like to show by brute force that, if $G=\mathbb{Z}_4$, $\tilde G=\mathbb{Z}_2\times \mathbb{Z}_2$ and $S=\{1,2,3,4\}$, then:
$$\forall \alpha,\beta =1,\dots,24, \exists i,j \in S \mid f_\alpha^{-1}(f_\alpha(i)+f_\alpha(j)) \ne \tilde f_\beta^{-1}(\tilde f_\beta(i)+\tilde f_\beta(j)) \tag 2$$
where $\{f_\alpha, \alpha=1,\dots,24\}=\operatorname{Sym}(S,G)$ and $\{\tilde f_\alpha, \alpha=1,\dots,24\}=\operatorname{Sym}(S,\tilde G)$, $\operatorname{Sym}(X,Y)$ being the set of bijections from $X$ to $Y$.
Is there any available fast resource to computationally check $(2)$?