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 allall 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
$$ \underline {\dot{h}} = \underline{M}_c -\underline\omega\times\underline h $$$$ \underline {\dot{h}} = -\underline{M}_c -\underline\omega\times\underline h $$
Hope this helps