Ideal operationnal amplifier with source
Will the voltage at A still be zero? Does it changes absolutely
  nothing to have Vs2 there?


To find the answer, write an equation for the node voltage \$v_A\$ using superposition.  For simplicity, assume the input resistance \$R_I\$ is infinite.

$$v_A = v_S \frac{R_2}{R_1 + R_2} + v_O \frac{R_1}{R_1 + R_2}$$

We also have

$$v_I = v_{S2}-v_A $$

$$v_O = Av_I $$

Thus, the first equation becomes

$$v_A = v_S \frac{R_2}{R_1 + R_2} + A(v_{S2}-v_A) \frac{R_1}{R_1 + R_2}$$

Rearranging yields

$$v_A =  \frac{v_SR_2 + Av_{S2}R_1}{(1 + A)R_1 + R_2}$$

This is the general expression for \$v_A\$.  As the gain \$A\$ gets very large,

$$v_A \approx \frac{Av_{S2}R_1}{AR_1} =  v_{S2}$$

In the ideal case where the gain is 'infinite', this expression is exact.  This is why we say there is a virtual short between the two input terminals of an ideal op-amp with negative feedback - the negative feedback ensures the inverting input voltage is identical to the non-inverting input voltage.