If $A$ is a nonempty set endowed with an associative binary operation, such that the left and right "translations" by any fixed $a\in A$ (say $\theta_a$ and $\gamma_a$, respectively) are bijections on $A$, then $A$ is a group (see e.g. this recent post).
If $A$ is finite, it's straightforward to infer the existence of left and right identities (later on proven to be equal) by the closure of the subsets $\Theta:=\{\theta_a, a\in A\}\subseteq \operatorname{Sym}(A)$ and $\Gamma:=\{\gamma_a, a\in A\}\subseteq \operatorname{Sym}(A)$, ensured in turn by operation's associativity; in fact: $\theta_a\theta_b=\theta_{ab}$ and $\gamma_a\gamma_b=\gamma_{ba}$ imply $\Theta,\Gamma\le\operatorname{Sym}(A)$, and thence $\exists\tilde e,\hat e\in A\mid\theta_{\tilde e}=\gamma_{\hat e}=Id_A$. If $A$ is infinite, in order to get the same outcome by using the same approach (subgroup criteria), I'd need to prove also the "closure by inverses", namely:
$$\forall a\in A, \exists\tilde a \in A\mid \theta_a^{-1}=\theta_{\tilde a} \tag 1$$
I can't figure out how to prove $ (1) $ without the notion of inverse of an element of $ A $, which is obviously not yet available at this stage of the process (existence of the identity). Where am I wrong?