Why moment equilibrium must be calculated around the center in this problem? Consider this problem, which ask for the value of the two normal forces (friction is negligible).

Easy, right? You calculate the moments around G, the center of mass, and equate them to 0 (the object slides without rolling), and get the left and right normal forces to be 359N and 664N, respectively. But why do you have to calculate the moment around the center of mass? 
As it turns out, if I calculate the moment around any other points and equate it to 0, the result will turn out wrong.
 A: Because the system is in equilibrium only in the CM frame. If you calculate the moment about any other point, you need to consider pseudo force as well, viz., 133.4N pointing to the left passing through the CM. Then you will get the correct answer again.
If you do not want to work in the CM frame, then you should notice that the angular momentum is
$${\bf L}={\bf r} _{CM} \times M{\bf v}_{CM}+{\bf L}_{CM}$$
Since the object is not rotating,
$${\bf L}_{CM}={\bf 0}$$
However, it is accelerating and hence ${\bf r}_{CM} \times M{\bf v}_{CM}$ is not a constant in general (unless ${\bf r}_{CM} // {\bf v}_{CM}$), and therefore, you cannot equate the moment to $0$.
A: Any force whose line of action is not through the centre of mass of a body can be transformed into the same magnitude and direction force acting through the centre of mass and a couple.
A couple (two parallel, non-colinear, equal in magnitude but opposite in direction force) has the special properties that its torque is independent of the point about which the torque is taken and produces only rotational motion.  
So each of forces $F, A$ and $B$ can be converted into a force through the centre of mass and the corresponding torque.

The middle diagram shows this done for force $F$ where the couple (in blue) exerts a torque $Fd$ clockwise and there is a force $F$ (in red) acting through the centre of mass.
So the right hand diagram shows a system of three couples (in blue) which you are asked to sum as zero and four forces acting at the centre of mass (in red and black) which do not sum to zero.
Since the sum of the four forces acting at the centre of mass is not zero they will exert a net torque about every point except the centre of mass.  
That is why the centre of mass is chosen for you to find the "moment equilibrium" because then the condition which is satisfied is that the body will undergo no rotational motion but only translational motion.
A: 
if I calculate the moment around any other points and equate it to 0, the result will turn out wrong.

By "points", you mean positions on the object.  But since there are unbalanced forces, the object is accelerating.  So these points are only at rest in a non-inertial frame.
Your choices are to either:


*

*Continue in the non-inertial frame, but add in a fictitious force that is the opposite of the acceleration.  When applied at the center of mass, the net torque will null out again.

*Stay in the non-inertial frame, but use the center of mass.  Since the fictitious inertial force acts at that point, the torque from it is zero and ignoring it doesn't matter.

*Switch to an inertial reference.  Then we see that the unbalanced torque is still present, but instead of causing a rotation, it causes the entire system to change its angular momentum about that reference point.

