If $G$ is a (possibly infinite) group, left and right multiplications establish the subgroups $\Theta:=\lbrace \theta_a \mid a \in G \rbrace \le \operatorname{Sym}(G)$ and $\Gamma:=\lbrace \gamma_a \mid a \in G \rbrace \le \operatorname{Sym}(G)$, such that $G \cong \Theta$ and $G \cong \Gamma$.
I'm wondering about the contrary: assume $G$ to be a (possibly infinite) set and suppose we can establish a pair of subgroups $\Theta:=\lbrace \theta_a \mid a \in G \rbrace \le \operatorname{Sym}(G)$ and $\Gamma:=\lbrace \gamma_a \mid a \in G \rbrace \le \operatorname{Sym}(G)$, such that:
i) $\theta_a(b)=\gamma_b(a)$ for all $a,b \in G$;
ii) $\theta_a\gamma_b=\gamma_b\theta_a$ for all $a,b \in G$.
Can we say that $\Theta$ and $\Gamma$ completely define some left and right multiplications, turning $G$ into some group? [i) would ensure the identity $ab=ab$, while ii) the associativity.]