BJT Differential Amplifier Common Mode & Differential Mode Gain
They simply do AC small-signal analysis. 

So you can skip \$V_{BE}\$ if you do AC analysis. 

The \$r_e\$ resistance "represents" the change in \$V_{BE}\$.

\$\Delta V_{BE} = i_e\cdot r_e\$ 

As for the voltage at point \$A\$. 

This voltage remains fixed due to the fact that we are again dealing with symmetrical AC signal (no AC current through R3) and "Imagine a symmetrical input signal wiggle in which input 1 rises by \$V_{IN}\$(a small signal variation) and input 2 drops by the same amount". 

For example, \$V_{IN}\$ if will increase \$I_{E1}\$ current from \$1mA\$ to let as say \$1.2mA\$ (due to \$V_{be1}\$ increase) and \$I_{E2}\$ will decrease by the same amount from \$1mA\$ to \$0.8mA\$

$$ΔIe1 = 1.2mA - 1mA = 0.2mA$$

$$ΔIe2 = 0.8mA - 1mA = - 0.2mA$$

So the AC current sum of the emitters currents \$Iee = ΔIe1+ΔIe2\$  will be equal to \$0A\$.

Because the AC component of a \$Ie1\$ and \$Ie2\$ are equal in magnitude but 180° out of phase.

This means that \$Iee\$ current is constant, no AC component. Hence the potage at point \$A\$ remains fixed.

(1.2mA + 0.8mA  = 2mA = constant).  



As for this "minus" sign in the gain equation. We usually omit this "minus" sign because we know what this "minus" sign represents/means. 
This "minus" only informs us that the output voltage is the 180-degree phase shift with respect to the input voltage.