This method is taken from Taylor's Classical Mechanics, in the "Two-Body Central-Force Problems" section. This goes way more in-depth than it needs to, and assumes uniform densities of the stars for ease of calculation (though this is not necessary).
tl:dr: The Lagrangian is independent of each star's angle off the center of mass, and independent of each star's rotational angle, so the total orbital and each star's rotational angular momentum are all independently conserved. This is achieved by the changing mass accounting for the change in rotational speeds. Note that this does not require tidal locking or circular orbits.
Take $\vec{R}$ to be the center of mass position.
$$
\vec{R}=\frac{m_1\dot{}\vec{r}_1+m_2\dot{}\vec{r}_1}{m_1+m_2}=\frac{m_1\dot{}\vec{r}_1+m_2\dot{}\vec{r}_2}{M}
$$
where $M≡m_1+m_2$.
We can subsequently define $\vec{r}_1=\vec{R}+\frac{m_2}{M}\vec{r}$ and $\vec{r}_2=\vec{R}-\frac{m_1}{M}\vec{r}$, where $\vec{r}=\vec{r}_1-\vec{r}_2$
The kinetic energy is
$$
T=\frac{1}{2}(m_1\dot{\vec{r}}^2_1+m_2\dot{\vec{r}}^2_2+\frac{2}{5}m_1 s_1^2\omega_1^2+\frac{2}{5}m_2 s_2^2\omega_2^2)
$$
where $s_i$ are the radii of the stars, and $\omega _i$ are the rotational velocities (not orbital). We can reduce this to
$$
T=\frac{1}{2}(m_1\dot{\vec{r}}^2_1+m_2\dot{\vec{r}}^2_2+\frac{2}{5}m_1 s_1^2\omega_1^2+\frac{2}{5}m_2 s_2^2\omega_2^2)
$$
$$
T=\frac{1}{2}[m_1(\dot{\vec{R}}+\frac{m_2}{M}\dot{\vec{r}})^2+m_2(\dot{\vec{R}}-\frac{m_1}{M}\dot{\vec{r}})^2]+\frac{1}{5}(m_1 s_1^2\omega_1^2+m_2 s_2^2\omega_2^2)
$$
$$
T=\frac{1}{2}[M\dot{\vec{R}}^2+\frac{m_1 m_2}{M}\dot{\vec{r}}^2]+\frac{1}{5}(m_1 s_1^2\omega_1^2+m_2 s_2^2\omega_2^2)
$$
This lets us define a new quantity, the reduced mass: $\mu≡\frac{m_1 m_2}{M}$. We finally get
$$
T=\frac{1}{2}M\dot{\vec{R}}^2+\frac{1}{2}\mu\dot{\vec{r}}^2+\frac{1}{5}(m_1 s_1^2\omega_1^2+m_2 s_2^2\omega_2^2)
$$
For the total Lagrangian, taking a potential energy $U=U(r)$, then, we obtain
$$
L=T-U=\frac{1}{2}M\dot{\vec{R}}^2+\frac{1}{2}\mu\dot{\vec{r}}^2+\frac{1}{5}(m_1 s_1^2\omega_1^2+m_2 s_2^2\omega_2^2)-U(r).
$$
We can see that since the Lagrangian is independent of $\vec{R}$, that $M\ddot{\vec{R}}=0$ or $\dot{\vec{R}}=const.$. This tells us total momentum is conserved, our first conservation law. This is because, in the closed system, $\dot{m_1}=-\dot{m_2}$, so $\dot{M}=0$.
Since $\dot{\vec{R}}=const.$, we can move into the CM rest frame, so $\dot{\vec{R}}=0$.
$$
L=\frac{1}{2}\mu\dot{\vec{r}}^2+\frac{1}{5}(m_1 s_1^2\omega_1^2+m_2 s_2^2\omega_2^2)-U(r).
$$
Let $\dot{\vec{r}}^2=\dot{r}^2+r^2\dot{\phi}^2$, $\omega_i ^2=\dot{\theta}_i^2$.
$$
L=\frac{1}{2}\mu(\dot{r}^2+r^2\dot{\phi}^2)+\frac{1}{5}(m_1 s_1^2\dot{\theta}_1^2+m_2 s_2^2\dot{\theta}_2^2)-U(r).
$$
The Lagrangian is independent of $\phi$, so we again obtain a conservation equation.
$$
\frac{d}{dt}\frac{\partial L}{\partial \dot{\phi}}=0
$$
$$
\frac{\partial L}{\partial \dot{\phi}}=\mu r^2\dot{\phi}=const=l_{orbit}
$$
This tells us that orbital angular momentum is conserved. Apparently
$$
\mu r^2 \ddot{\phi} + 2 \mu r \dot{r} \dot{\phi} + \dot{\mu} r^2 \dot{\phi} = 0.
$$
If you were curious, $\dot{\mu}=\frac{\dot{m}_2 (m_1-m_2)}{M}=\frac{\dot{m}_1 (m_2-m_1)}{M}$.
Again, since the Lagrangian is independent of both $\theta _1$ and $\theta _2$, each star's individual rotational angular momentum is conserved.
We can find that
$$
\frac{2}{5}m_i s_i ^2 \dot{\theta}_i = const = l_{rot,i}
$$
and apparently
$$
m_i s_i ^2 \ddot{\theta}_i + 2 m_i s_i \dot{s_i} \dot{\theta}_i + \dot{m}_i s_i ^2 \dot{\theta}_i = 0,
$$
giving us our last constraint equation. Solving the Lagrangian for $r$ tells us the equations of motion.
Best Answer
Of course you can define such a quantity, but the question is: does it mean anything physically?
Contrary to what has been stated in some of the answers/comments, this quantity is not comparable to a "normalized" dipole moment. A dipole is a system of two charges equal in magnitude but opposite in sign. The corresponding dipole moment, which is of great importance for the description of many phenomena in electromagnetism, is defined as the product of the magnitude of one of the charges and their displacement vector. It would be equal to the numerator of your expression if you put the negative charge to the origin, the only term in the sum would be the product of the charge and the position (displacement) vector. So far so good: but what about the statement that your expression would then be a "normalized dipole moment"? Let us look at the denominator: in case of a dipole, the sum of both charges amounts to zero, so we would have a division by zero. This does not give us something normalized, but rather something ill-defined. Hence, this concept does not make much sense.
This problem remains for any system where the sum of charges is equal to zero, i.e. for neutral systems. Thus, it is not defined for many physically important situations and even in cases where it is, it does not tell us anything about the properties of that system.
Your "charge momentum" is related to the current density, which is given by the product of charge density and velocity. Its time derivative would simply be the rate of change of a current, and meaningful for example in a system with time-dependent electrostatic potential: this can be found in electrotechnical application when dealing with alternate currents. But can this be compared to a force?
To answer this question, we have to examine the nature of the comparison of the quantity $q\vec{v}$ to momentum in mechanics. What makes momentum so special is the fact that it is conserved for closed systems, i.e. systems without any external forces. But what is required in order for a certain quantity to be conserved?
One of the key principles of classical mechanics, Noether's theorem, tells us that conserved quantities (also called "conserved charges" or "Noether charges") are related to continuous symmetries of a system (there are various sources on the internet and books which describe this principle in as much detail as one might imagine). The conservation of momentum is a consequence of translational invariance of a physical system. In order to be comparable to momentum, your "charge momentum" would have to correspond to a continuous symmetry of the underlying system, but it turns out that there is none.
However, it is not completely unrelated to that concept either. While your "charge momentum" is no conserved quantity, charge itself is indeed one, corresponding to the $U(1)$ symmetry of electromagnetism. Within the framework of Noether's theorem, there also exists the notion of a so-called conserved four-current $J^\mu$, which has to satisfy
$$\partial_\mu J^\mu=0,$$
i.e. its four-divergence vanishes. In the case of the $U(1)$ symmetry, splitting space and time components and writing out the equation explicitely gives
$$\frac{\partial\rho}{\partial t}+\partial_i j^i=0,$$
which is nothing but the continuity equation where $\rho$ is charge density and $j^i$ is three-dimensional current density, which is related to your "charge momentum", as was already pointed out in the comments. There is no indication that the latter is conserved by itself.
Since there is no conservation of "charge momentum", i.e. no analogue to Newton's first law, applicability of an analogue of Newton's second law, which states that force is defined as the time-derivative of conserved momentum is highly doubtful. Furthermore, there is no reason to assume that the principle of "actio=reactio" (Newton's third law) should hold.