Power factor calculation of an offset sine wave across a non-ideal capacitor
In your example, the total loss dissipated in heat corresponds to the sum of individual losses in the resistors. You correctly calculate them with \$P_R=I_{rms}^2R\$ with \$R\$ the considered resistance and \$I\$ the current in the resistor. I simulated the below sketch with a 100-\$\Omega\$ resistance across the capacitor. The source delivers 10 V peak at a 1-kHz frequency with a 5-V offset.


You compute the average power - or the real power - by multiplying \$i_{in}(t)\$ and \$i_{tot}(t)\$ and average the result across a cycle. This always works regardless of the type of waveform or distortion. If you deal with sinusoidal signals, then you can involve \$cos\phi\$ applied to the ac component as you mentioned and deal with rms voltage and current. 

The below simulation results show the instantaneous waveforms:



If I average the instantaneous power which is plotted in the low-side window, the simulator gives me 744 mW. The rms current flowing in resistor \$R_1\$ is 85.74 mA while in the 1-\$\Omega\$ resistance it is 96.4 mA: \$P_{avg}=85.74m^2\times 100 + 96.4m^2\times 1 = 735.1\;\rm mW + 9.3\;\rm mW = 744.4\;\rm mW\$ in line with the measurement (these numbers should be perfectly equal but you need to exactly position the measurement cursors to isolate a period). 

If we now measure the rms ac components (without the dc offset) of the input current and voltage, we find \$V_{in,rms}=7.07\rm\;V\$ and \$I_{tot,rms}=82.7\rm\;mA\$. The phase shift is 31.6° which leads to an intermediate real power of \$V_{ac,rms}I_{ac,rms}\rm cos(\phi)= 498\rm\; mW\$. The dc current is the offset (5 V) divided by the total resistance (\$101\rm\; \Omega\$) and it amounts to 50 mA or 248 mW dissipated in the resistance. Sum 498 mW to 248 mW and you obtain 746 mW in line with what we found.