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 & S bands only. In the Q band $\Delta J = 0$ between vibrational levels n to $n+1$ and is $-2$ and $+2$ for O & S branches respectively. The Raman selection rules are $\Delta n = \pm 1$ and $\Delta J = 0, \pm 2 $.
If you look at a rotation-vibration energy level diagram for IR transitions you should by adapting this be able to work out your values for your Raman transitions.
There is a figure showing energy levels at
The Q band transition ($\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.