Why does the Born equation give the Gibbs free energy of solvation rather than enthalpy of solvation?
1) Because $-\Delta G$ is the maximum possible nonexpansion work done by a system in a constant temperature and pressure process. You can write $$\Delta G \le w_{non PV}\quad (\text{constant T and P, closed system})$$
where the equality sign holds for reversible changes.

Here's proof:
$$\begin{align*}
H &amp;= U + P V&amp;\text{(definition of enthalpy)}\\
dH &amp;= dq + dw + d(PV) &amp;\text{(because dU = dq + dw)}\\
dG &amp;= dH - T dS - S dT &amp;\text{(because G = H - TS)}\\
&amp;=dq + dw + d(PV) - T dS -S dT &amp;\\
&amp;= dq + dw + d(PV) - T dS &amp; \text{(when the change is isothermal)}\\
&amp;= dw_{rev} + d(PV) &amp;\text{(if reversible,}\ dq_{rev} = T dS\text{)}\\
&amp;= dw_{electrical,rev} + dw_{expansion,rev} + P dV + V dP &amp; \text{(partitioning the work)}\\
&amp;= dw_{electrical,rev} + V dP &amp; (dw_{expansion,rev} = - P dV)\\
&amp;= dw_{electrical,rev} &amp; \text{(if pressure is constant)}\\
\end{align*}$$

2) It isn't the scaling to one mole that makes the result a Gibbs free energy (see 1); that just makes it a molar Gibbs free energy. You can use either extensive or intensive Gibbs free energies in calculations; whatever is most convenient. Just make it clear (either by context or notation) which one you're using.