Wing pitching moment decomposition into two terms Summary: Assuming constant angle of attack $\alpha$, speed $V$, and lift $L$ applied at the center of pressure CP (whose position is $x_{CP}$), the distance $(x_{CP} - x_O)$ between the moment reference point $O$ (whose position is $x_O$) and $CP$ determines the sign (positive, negative or zero) of the pitching moment as long as $L\neq 0$:

$$M_O = (x_{CP} - x_O)L$$                   

However, the moment $M_O$ depends on $\alpha$, i.e. $M_O=M_O(\alpha)$, since changing $\alpha$ while keeping the reference point $O$ position constant, varies the magnitude $M_O$ since both lift $L$ and the position $x_{CP}$ vary.  

Interestingly, when $O=AC$ and $x_O=x_{AC}$, the moment does not change with changing $\alpha$:

$$ M_{O}(\alpha) = M_{AC} = (x_{CP} - x_{AC})L = constant$$

We can move lift $L$ from its application point $x_{CP}$ to point $x_{AC}$ while adding a constant free pitching $M_{AC}$ whose magnitude is $(x_{CP}-x_{AC})L$. The moment $M_{AC}=0$ for symmetric wings and $M_{AC}\neq0$ for cambered wings. 

In the case of a cambered wing, the equation 
$$ M_{AC} = (x_{CP} - x_{AC})L$$ predicts that $$M_{AC}=0$$ when lift $L=0$. However, we know that a cambered wing has a a constant nonzero moment $M_{AC}$ for any $\alpha$ even when $L=0$ at the zero-lift $\alpha_0$. How do we transform the equation for $M_{AC}$  so it becomes equal to the sum of two terms, one solely due to camber and one solely due to lift:
$$ M_{O}(\alpha) = M_{AC} = (x_{CP} - x_{AC})L= M_{camber}+M_{lift}$$

where the term $M_{camber}\neq 0$ for any $\alpha$?
Interestingly, when $O=AC$ and $x_O=x_{AC}$, the moment does not change with changing $α$


Yes, but only if your lift is also applied at the AC! That's the whole point of the aerodynamic center. You replace the pressure distribution with a lift force (actually a resultant force, but let's neglect drag here) and a moment. If you place the lift force at the center of pressure, then the resultant moment about that point is zero. Then, the moment about any other point $O$ is only "created" by the resultant force. So in that case indeed: 
$$M_O = (x_{CP} - x_O)L$$

If you now place the lift at your AC and also O=AC, then:

$$M_O = M_{AC}$$

This $M_{AC}$ is independent of $\alpha$. If you now move to any different O, you get:
$$M_O = M_{AC} + L (x_O - x_{AC})$$

The first term $M_{AC}$ is what you are calling "$M_{camber}$" and the second term is your "$M_{lift}$".