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. $\rho_{\{e\}} = \Delta$ (the equality relation)
5. $\emptyset = \rho_{\emptyset}$
6. $G\times G=\rho_G$
7. $(\rho_A)^{\cup} = \rho_{A^{\cup}}$
8. $(G\times G)\setminus\rho_A = \rho_{G\setminus A}$.

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

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. $\rho_{\{e\}} = \Delta$ (the equality relation)
5. $\emptyset = \rho_{\emptyset}$
6. $G\times G=\rho_G$
7. $(\rho_A)^{\cup} = \rho_{A^{\cup}}$
8. $(G\times G)\setminus\rho_A = \rho_{G\setminus A}$.
9. $A=B$ iff $\rho_A=\rho_B$.

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

Concerning the original question about why the axioms of RA are ``the right ones'', let me
add a pointer to the recent paper

Rob Egrot and Robin Hirsch
First-order axiomatisations of representable relation algebras need formulas of unbounded quantifier depth
The Journal of Symbolic Logic. 2022;87(3):1283-1300

On the second page of this article the authors write:

So Tarski’s axiomatisation of RA was in a sense a failure, but it also turns out to be a remarkable success, for the following reason. Relation algebra equations correspond exactly with first-order statements about binary relations that can be stated using at most 3 variables (see [25, theorems 3.9(viii)(ix)] for a proof), and, moreover, a relation algebra equation is valid in RA if and only if the corresponding first-order sentence is provable (in classical proof systems) using at most 4 variables [20, theorem 24]. So Tarski’s axioms neatly capture an intuitively meaningful fragment of the first-order theory of binary relations. Furthermore, an axiomatisation such that the ‘4’ in the statement above could be replaced by ‘5’ would require an infinite number of additional axioms (this result seems to lack a precise statement in the literature, but it can be pieced together from the material in e.g. [12, section 6]).

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''.

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. $\rho_{\{e\}} = \Delta$ (the equality relation)
5. $\emptyset = \rho_{\emptyset}$
6. $G\times G=\rho_G$
7. $(\rho_A)^{\cup} = \rho_{A^{\cup}}$
8. $(G\times G)\setminus\rho_A = \rho_{G\setminus A}$.

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