Torque on rolling cylinder Consider a rod with two equal forces acting parallel to the direction of velocity of a rolling rod, one at each end. When analysing problems involving torque, the choice of axis should not matter. However, when we consider the axis as one end of the rod, there seems to be an unbalanced torque at the other end which will cause angular acceleration. However, in reality there is no such acceleration. What causes this discrepancy?
Note: consider 2nd example in image with F1 = F2 and cylinder rolling along long end

 A: 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 $\tau=a\times F_1 + b \times F_2$. 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 $\tau=a \times F+b\times F=(a+b)\times F=(\vec{OA}+\vec{OB})\times F$. 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.
A: Even if the object is accelerating you can take the torques about the center of mass (CM) considering only the forces in the inertial frame; using any point in the object other than the CM for evaluating torques, you must consider the non-inertial forces in the accelerating frame of reference.  See the link provided by@BowlOfRed in a comment above.  See the text Mechanics, by Symon for more details.
A: When summing torque, the choice of axis only does not matter when the net force is zero. So only in statics this is true.
Otherwise you transform torque from one place to another with $$ \boldsymbol{\tau}_A = \boldsymbol{\tau}_B + \boldsymbol{r}_{B/A} \times \boldsymbol{F}$$
Specifically in dynamics it is convenient to sum torques about the center of mass, because this decouples the equations of motion.
The center of mass has the property of decoupling momentum in translational and rotational parts
$$ \begin{aligned} 
\boldsymbol{p} & = m \boldsymbol{v}_C \\
\boldsymbol{L}_C & = {\rm I}_C \boldsymbol{\omega} \\
\end{aligned} $$
and by application of Newton's 2nd law the equations of motion are also decoupled
$$ \begin{aligned} 
\boldsymbol{F} & = \tfrac{\rm d}{{\rm d}t} \boldsymbol{p} = m \boldsymbol{a}_C \\
\boldsymbol{\tau}_C & = \tfrac{\rm d}{{\rm d}t} \boldsymbol{L}_C = {\rm I}_C \boldsymbol{\alpha} + \boldsymbol{\omega} \times {\rm I}_C \boldsymbol{\omega} \end{aligned} $$
A typical example of this point is when hanging a rod from the ceiling using two equal strings at its ends, and you cut one of the strings. You will find the tension on the other string only if you realize that the sum of forces is not zero  and the torque evaluated should consider that.
