Gibbs energy derivative vs equilibrium for aqueous reaction
To derive the Gibbs' free energy of an ideal solution you would invoke Raoult's law, which says that 
$$\mu_i(l) = \mu_i^*(l)+RT\ln\left(\frac{p_i}{p*}\right)= \mu_i^*(l)+RT\ln\chi_i$$ 
Here the asterisk refers to the pure substance. Using the notation in the OP, the total number of moles of reagent and product in solution is

$$n_{tot}=2(n_{ini}-\xi)+\xi+n_{solv}=2n_{ini}-\xi+n_{solv}$$

where $n_{ini}$ is the initial amount of A or B (assumed equal), so that one can write for reagents A and B:
$$\mu_i(l) = \mu_i^*(l)+RT\ln\left(\frac{n_{ini}-\xi}{2n_{ini}-\xi+n_{solv}}\right)$$ 
and for product C:
$$\mu_C(l) = \mu_C^*(l)+RT\ln\left(\frac{\xi}{2n_{ini}-\xi+n_{solv}}\right)$$ 
There is also a term for the solvent:
$$\mu_{solv}(l) = \mu_{solv}^*(l)+RT\ln\left(\frac{n_{solv}}{2n_{ini}-\xi+n_{solv}}\right)$$ 
which I ignored in earlier edits. 
Adding terms to compute the total Gibbs energy and taking the derivative wrt the extent of reaction $\xi$ then results in 

$$\begin{align} \Delta G' &amp;= \Delta \mu^* + RT \ln\left(\frac{\xi(2n_{ini}-\xi+n_{solv})}{(n_{ini}-\xi)^2}\right)  \\  &amp;= \Delta \mu^* + RT \ln\left(\frac{n_Cn_{tot}}{n_A n_B}\right)  \\&amp;= \Delta \mu^* + RT \ln \left(\frac{\chi_C}{\chi_A \chi_B}\right) \end{align} $$ 

($^\dagger$)

Note that here the standard state for each substance refers to the pure liquid in equilibrium with its vapor at the same temperature. To obtain an expression relating a change in standard free energy to the concentration equilibrium constant requires expressing mole fractions in terms of concentrations. This is generally done for dilute solutions where $n_{solv} \approx n_{tot}$ by writing $\chi_i\approx c_iM_{solv}/\rho_{solv}$, which leads to 

$$\Delta G' = \Delta \mu^* - RT \ln \left( \frac{M_{solv}}{\rho_{solv}} \right) + RT \ln \left(\frac{c_C}{c_A c_B}\right)  $$ 
Then by redefining the standard states as
$$\mu_i^\circ = \mu_i^* + RT \ln \left( \frac{M_{solv}}{\rho_{solv}} \right) $$ leads to $$\Delta G' = \Delta \mu^\circ + RT \ln \left(\frac{c_C}{c_A c_B}\right)  $$ 
If you think the standard state looks strange, consider that if $\chi_i = c_iM_{solv}/\rho_{solv}$ then $$\chi_i(\pu{c^\circ_i= 1 M}) = c_i^\circ \frac{M_{solv}} {\rho_{solv}}$$ and so we can write  

$$\mu_i^\circ = \mu_i^* + RT \ln \left( \chi_i(\pu{c^\circ_i= 1 M}) \right) $$ which is more in line with canonical definitions of standard states. The new standard state is defined as a 1 M ideal solution with chemical potential related to that of the pure solute by the above equation.  



$^\dagger$/TL;DR :  The mysterious extra $RT$ term in the OP is gone. It's important to keep a term for the solvent. And a lesson from these repeated edits: careful doing math when tired!