Raman Spectroscopy: How to calculate their constants?
In Raman spectroscopy the difference in frequency between the exciting line and the measured signal frequencies contains the information you want. These values, made positive if necessary, give the ro-vibrational transition frequencies.

You are assuming that the molecule is vibrating with an anharmonic potential, such as a Morse potential. The energy levels are given by $E_n= \omega_e(n+1/2) - x_e\omega_e(n+1/2)^2$ (in wavenumber units) with quantum number $ n = 0, 1, ...~$, and $\omega_e$ the frequency and $x_e$ is the dimensionless anharmonicity. 

In addition to the vibrational energy there is rotational energy whose levels are $E_J =B_eJ(J+1)$ for rotational constant $B_e$ (in wavenumbers) and quantum number $J=0, 1, 2,...$ and ignoring centrifugal distortion. You will also have to use different $B_e$ values for the $n=0$ and $n=1$ vibrational levels since the bond length is slightly different between (anharmonic) vibrational levels.

You know that the rotational-vibrational levels appear in groups  O, P, Q, R, S in increasing frequency and that Raman has O, Q &amp; S bands only. In the Q band $\Delta J = 0$ between vibrational levels n to $n+1$ and is $-2$ and $+2$ for O  &amp; S branches respectively. The Raman selection rules are $\Delta n = \pm 1$ and $\Delta J = 0, \pm 2 $. 

If you look at a rotation-vibration energy level diagram for IR transitions you should by adapting this be able to work out your values for your Raman transitions.

There is a figure showing energy levels at
 https://en.wikipedia.org/wiki/Rotational%E2%80%93vibrational_spectroscopy

The Q band transition ($\Delta n =0, \Delta J =0$) in your notation is $\omega_0 = \omega_e - 2x_e \omega _e $ and the other branches calculated similarly with $\Delta J = \pm 2$ as appropriate.