Electronic – Why do we add Resistors on differential op. amp

1) Layout is critical, decoupling caps are required, watch the common mode input range on the amplifier, though if you are using +/- supplies you will likely be OK.

2) Since you are measuring across a low impedance shunt (presumably) the input impedance of the amplifier will be much larger than the shunt impedance so it should have little effect.

3) As long as the input resistors are >> than the shunt impedance you don't have an issue.

As Null mentioned there are dedicated current shunt amplifier that perform this function.

Brief Background
Suppose you have a linear network which has two input ports with input voltages \$V_1\$ and \$V_2\$ as shown in figure below:

Then, since \$V_1 = \frac{V_1-V_2}{2}+\frac{V_1+V_2}{2}\$ and \$V_2=\frac{V_2-V_1}{2}+\frac{V_1+V_2}{2}\$. Thus we have:

Then you can transform the circuit as shown below:

Here the common mode voltage is: \$V_{cm} = \frac{V_1+V_2}{2}\$ and the differential voltage is: \$\frac{V_{diff}}{2} = \frac{V_1-V_2}{2}\$. Since the circuit is linear, superposition is valid. So we can say that the total response will be sum of these two.

The first one is the the common-mode circuit and the second one is the differential circuit. Here you can use all the tricks for the differential half and the common-mode half which you may know.
Your Example
The complete circuit for the example you provided will be:

Here the two inputs are: \$V_1=V_{cm}+V_{in}\$ and \$V_2 = V_{cm}\$.
If you use superposition here with \$V_{cm}=0\$, you get the circuit which you have shown in your question. This is the differential part of the circuit.
If you instead make \$V_{in}=0\$, you get the common-mode circuit:

I leave it to you now to analyze it.