Gibbs free energy where T, V, and P change simultaneously
Your original equation is incorrect.  It should read $\Delta G=\Delta H -\Delta (TS)$, and $\Delta (TS)\neq T\Delta S+S\Delta T$ unless the changes are infinitesimal.  

That said, for an ideal gas, irrespective of path, $$\Delta S=S_f-S_i=C_p\ln{(T_f/T_i)}-R\ln{(P_f/P_i)}=C_v\ln{(T_f/T_i)}+R\ln{(V_f/V_i)}$$and$$\Delta H=C_p(T_f-T_i)$$So, $$\Delta G=G_f-G_i=C_p\Delta T-\frac{(T_f+T_i)}{2}\Delta S-\frac{(S_f+S_i)}{2}\Delta T$$Clearly the 3rd term on the rhs depends on either $S_i$, $S_f$, or both, so $\Delta G$ depends on these.  It then all comes down to how you assign your reference state values for G, H, and S, such that $G_{ref}=H_{ref}-T_{ref}S_{ref}$.