Torque on rolling cylinder

Linking this to a (somewhat) similar question I have asked before; Confusion about force analysis in 3D space

Let us take the point through which $F_2(=F_1=F)$ acts in the second figure as the origin O, and let the length of the rod be $L$. Assume the rod is uniform with mass M. Now let us denote the direction of the rod and the direction of the forces as the positive $i-$ and $j-$ vectors respectively. We see that the net force on the system is $2Fj$ and the torque on the system is $sum r_i times F_i=Li times Fj=LF k$. Now we see that $LFk cdot 2Fj=0$, so the system simplifies to a single force acting through the COM of the rod with no net torque.

In regards to your statement that the choice of axis should not matter:

Assume we have two forces, $F_1$ and $F_2$, acting at points with position vectors $a(=vec{OA})$ and $b(=vec{OB})$ relative to some origin O respectively. Now the torque about the system is $tau=atimes F_1 + b times F_2$. Now if this system is a couple, $F_1=-F_2=F$ so $tau= a times F -b times F=(a-b)times F=vec{BA} times F$. This is why the torque of a couple does not depend on the point chosen to be the origin, since $vec{BA}$ is the position vector of one point relative to the other,and O is not featured in the equation. However, in this case, with $F_1=F_2$, we have $tau=a times F+btimes F=(a+b)times F=(vec{OA}+vec{OB})times F$. Now we see that unlike previously, the torque of this system is actually dependent on the point chosen to be the position vector. However, as I have shown above, this system of forces simplifies to only a translational force; in fact, although I did it for a specific point, I believe it can be generalised to show that no matter what point you take as the origin, the system simplifies to a single force acting through the COM of the rod.