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
First, angular momentum isn't measured about an axis. It's measured about a point.
Second, well, of course the angular momenta about different points will be different in general. But they will each be conserved -- there's no need for the point to be in the axis of rotation or even in the same galaxy as the rotating object you're interested in.
Now, about your example. The total angular momentum, on the three block system, is definitely conserved. Why wouldn't it be? What you're perhaps missing is the fact that there will be in general an angular momentum associated with the 3rd block as well, even if it's traveling on a straight line. For example, say you have a block with momentum $\vec{p}$ traveling in a straight line at constant velocity until it hits the block at one end of the string (at the point of closest approach to the center of mass). Say the string has length $2\ell$. Now let's calculate the angular momentum of this third block about the center of mass of the two rotating blocks when it's at a distance $r$ from that center of mass.
$$\vec{L} = \vec{r} \times \vec{p} = rp \sin \theta \hat{z} = rp \frac{\ell}{r}\hat{z} = rp \hat{z} $$
(The $\hat{z}$ is just my choice in how to orient the system in space).
Notice now that $L$ does not depend on the distance $\vec{r}$. It's a constant of the motion, as promised, so long as $\vec{p}$ itself is conserved. Notice that there was nothing special in this derivation about the center of mass: it could've been literally any point and the conclusion still would be valid.
Now you can work out what will happen in the collision, assuming that linear momentum is conserved, and you'll prove explicitly that angular momentum is conserved in this process.