Does total angular momentum of the Earth-Moon system include individual rotational angular momenta? To calculate the angular momentum of a body we need to specify a point (or an axis?) from which to define the displacement vector $\vec{r}$, so that $\vec{L} = \vec{r} \times \vec{p}$.
For a rigid body, the formula becomes $\vec{L}=I\vec{\omega}$, assuming the moment of inertia is not a tensor. So in this case we need to specify an axis.
NOW: what if I wanted to calculate the total angular momentum of the Earth-Moon system?
What is the most sensible point/axis to choose? 
I would say, intuitively, the centre of mass of the system.
So about the CoM,
$$\begin{align}\vec{L}_\text{tot}
&= \vec{L}\text{ of the Earth due to its orbit about centre of mass} \\
&+ \vec{L}\text{ of Moon due to its orbit about centre of mass} \\
&+ \text{angular momenta due to rotations of Earth and Moon around their own axes}?\end{align}$$
I am not sure if and how to include the rotations of Moon and Earth around their axes: I know they have to enter somehow because of the tidal friction effect on the orbit of the Moon, but I don't know how to reconcile this with the fact that I chose the centre of mass as the point of reference here, and none of the rotation axes have anything to do with the centre of mass of the Earth-Moon system.
 A: Consider two bodies A and B. With respect to an inertial coordinate system with origin at point O, the coords of the particles in A are vectors $x_{a}\in V_{3}$ with $a=1,2,\ldots, N_{A}$ and similarly the coords of the particles of B are $x_{b}\in V_{3}$ with $b=1,2,\ldots,N_{B}$.  The momenta wrt the inertial frame with origin at point O of the particles of A are $p_{a}\in V_{3}$ and the momenta of the particles of B are $p_{b}\in V_{3}$. The total angular momentum of the system wrt point O is,
$$
J=\sum_{a}x_{a}\times p_{a}+\sum_{b}x_{b}\times p_{b} \  .
$$
Let's introduce the centre of mass of body A as $X_{A}\in V_{3}$,
$$
X_{A}=\frac{\sum_{a}m_{a}x_{a}}{\sum_{a}m_{a}}=\frac{\sum_{a}m_{a}x_{a}}{M_{A}}
$$
and the centre of mass of body B as,
$$
X_{B}=\frac{\sum_{b}m_{b}x_{b}}{\sum_{b}m_{b}}=\frac{\sum_{b}m_{b}x_{b}}{M_{B}}
$$
Adding and subtracting the centre of mass coords,
$$
J=\sum_{a}(x_{a}-X_{A})\times p_{a}+\sum_{b}(x_{b}-X_{B})\times p_{b}+X_{A}\times \sum_{a}p_{a}+X_{B}\times \sum_{b}p_{b} \ .
$$
The first two terms on the RHS are the angular momenta of the bodies about their respective centres of mass $X_{A}$ and $X_{B}$. Let's write these contributions as $J_{A}\in V_{3}$ and $J_{B}\in V_{3}$. The total angular momentum is now,
$$
J=J_{A}+J_{B}+X_{A}\times \sum_{a}p_{a}+X_{B}\times \sum_{b}p_{b} \ .
$$
Let the linear momentum of the particles of body A be,
$$
P_{A}=\sum_{a}p_{a}
$$
with a similar formula for the sum of the momenta of the particles of body B. Put these formulae into the equation for the total angular momentum,
$$
J=J_{A}+J_{B}+X_{A}\times P_{A}+X_{B}\times P_{B} \ .
$$
This is precisely the splitting of the total angular momentum that Harold wrote in his question. $J_{A}$ and $J_{B}$ are the angular momenta of A and B about their own centres of mass. $X_{A}\times P_{A}$ is the orbital angular momentum of A about the origin O of the inertial coords and $X_{B}\times P_{B}$ is the orbital angular momentum of B about O. The point O could be taken as the centre of mass of the complete system, but it doesn't matter for the above result.
