Given that:

* Euler equations are general, there exist no special or modified forms. Simply some terms are equal to zero if some conditions are satisfied
* I do not quite understand what $S_{ij}$ is in your expression 

Anyway, if the centre of mass of the body coincides with the origin of the body frame, the total angular momentum of the satellite can be written as
$$
\underline{H} = J\underline{\omega} + \underline{h}
$$ 

Where $J$ is the inertia tensor, $\underline{\omega}$ is the angular velocity vector, $\underline{h}$ is the total angular momentum of ***all*** the rotating internal equipments. The second law of motion states that the derivative of the total angular momentum equals the sum of the external torques

$$
\frac{D\underline{H}}{Dt}=\underline{\dot H}+\underline{\omega}\times\underline{H}=\underline{M}
$$

Neglecting the variation of the inertia tensor, the previous can be written as
$$
J\underline{\dot \omega}+\underline\omega\times J\underline\omega + \underline {\dot{h}} +\underline\omega\times\underline h=\underline M
$$
If you have 3 rotors each aligned with one of the satellite axes, then $\underline h$ is
$$
\underline h = 
\begin{bmatrix}
J_{r_1}\omega_{r_1} & J_{r_2}\omega_{r_2} & J_{r_3}\omega_{r_3}
\end{bmatrix}^\rm T
$$
You can name the vector of control torques (coming from some control technique as you mentioned) as
$$
\underline{M}_c = -\underline {\dot{h}} -\underline\omega\times\underline h
$$

and compute the needed torque each wheel needs to provide

$$
\underline {\dot{h}} = \underline{M}_c -\underline\omega\times\underline h
$$

Hope this helps