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There is one natural candidate that leaps to mind, namely the construal of $2^G$ as a relation algebra for each group $G$ $\ldots$

This construction will not answer your question, since each relation algebra of the form $2^G$ is representable. To see this, for each $A\subseteq G$ define a binary relation $\rho_A$ on $G$ by $ \rho_A :=\{(x,y)\in G^2\;|\; x^{-1}y\in A\}. $ Now check that

1. $\rho_A\cup \rho_B = \rho_{A\cup B}$
2. $\rho_A\cap \rho_B = \rho_{A\cap B}$
3. $\rho_A\circ \rho_B = \rho_{A\cdot B}$
4. $\emptyset = \rho_{\emptyset}$
5. $G\times G=\rho_G$
6. $(\rho_A)^{\cup} = \rho_{A^{\cup}}$
7. $(G\times G)\setminus\rho_A = \rho_{G\setminus A}$.

These show that $A\mapsto \rho_A$ is a representation of $2^G$. Thus the $2^G$-construction is ``insufficiently surjective''.