Exercise 2.2.2 of An Invitation to Universal Algebra says:
If $G$ is a group, let us define an operation $\delta_G$ on $|G|$ by $\delta_G(x, y) = x · y^{−1}$. Does the pair $G' = (|G|, \delta_G)$ determine the group $(|G|, ·, {}^{−1}, e)$? (I.e., if $G_1$ and $G_2$ yield the same pair, $G'_1 = G'_2$, must $G_1 = G_2$? Some students have asked whether by “=” I here mean “$\cong$”. No, I mean “=”.)
So far I have been able to find an algorithm to recover $G$ from $G'$ (if one exists, which I doubt). I did try the other way around for some simple examples. For $\mathbb{Z}_2$, the group is unaffected by the $\delta$ operation since both members are already their own inverse. After applying the $\delta$ operation to $\mathbb{Z}_3$, there is no right identity, so it isn’t even a group. Thank you for any advice.
The group $G=(|G|,\cdot,x^{-1},e)$ is recoverable from $G':=(|G|,\delta_G)$ when $\delta_G(x,y):=x\cdot y^{-1}$.
To recover the identity $e$ of $G$ from $\delta_G$, observe that $e$ is the unique element in the image of the function $\delta_G(x,x)$.
Once you have $e$, recover the inverse using the formula $x^{-1} = \delta_G(e,x)$.
Once you have the inverse, recover the multiplication using the formula $x\cdot y = \delta_G(x,y^{-1})$.
In particular, if $G_1'=G_2'$, then $|G_1|=|G_2|$ and $\delta_{G_1}=\delta_{G_2}$. Recovering the identity element by the above formula leads to $e_1=e_2$. Then recover inverse and multiplication to get ${x}^{-1_1}={x}^{-1_2}$ and $x\cdot_1 y = x\cdot_2 y$, so $G_1=(|G_1|,\cdot_1,x^{-1_1},e_1)=(|G_2|,\cdot_2,x^{-1_2},e_2)=G_2$.