newtonian mechanics - Conservation of spin angular momentum in a close binary system

Consider a simple model of a close stellar binary, of mass $m_1$ and $m_2 < m_1$, moving on circular orbits around the system's barycenter (no eccentricity, to simplify things). Both star's rotation is permanently tidal locked. The total angular momentum relative to the barycenter is conserved : \begin{equation}\tag{1} \vec{L} = \vec{L}_1 + \vec{L}_2 + \vec{S}_1 + \vec{S}_2 = \textrm{constant}, \end{equation} where $\vec{S}_i$ is the spin of a star (i.e. its angular momentum around its own center of mass). These vectors are all aligned. Using Newton's theory of gravitation, we can prove that \begin{equation}\tag{2} L_{\text{orbital}} = || \vec{L}_1 + \vec{L}_2 || = \frac{m_1 \, m_2}{m_1 + m_2} \, \sqrt{G (m_1 + m_2) a}, \end{equation} where $a = r_1 + r_2$ is the distance between both stars. Lets write $M \equiv m_1 + m_2$ to simplify. Also, since the stars are tidal locked ; $\omega_1 = \omega_2 = \omega_{\text{orbital}} \equiv \omega$ : \begin{equation}\tag{3} S_{\text{tot}} = || \vec{S}_1 + \vec{S}_2 || = (I_1 + I_2) \, \omega, \end{equation} where $I_i$ is the inertia moment of a star around its center. If the stars are approximately spherical, then $I_i = \alpha_i \, m_i \, R_i^2$, where $\alpha_i \approx \frac{2}{5}$ (or any number smaller than 1). The orbital angular velocity is \begin{equation}\tag{4} \omega = \sqrt{\frac{GM}{a^3}}. \end{equation}

Now, suppose that the stars are allowed to exchange some matter: $m_1$, $m_2$, $a$, $\omega$, $I_1$ and $I_2$ are now functions of time ($M$ is conserved, though).

If the whole system is isolated, the total angular momentum (1) is conserved.

How can we justify that both $L_{\text{orbital}}$ and $S_{\text{tot}}$ are separately conserved (maybe approximately) ?