Finding Thevenin Resistance Amp Op
A common approach when trying to find the Thevenin equivalent looking into a certain node or between two nodes (as in this case), is to place an ideal current source, say \$I_T\$ in parallel with the node(s) of interest and then finding the ratio of the voltage created by this current source \$V_T\$ and such current. $$R_T = \dfrac{V_T}{I_T}$$ 
Now back to your current circuit, you've followed the steps correctly by replacing \$V_{IN}\$ with it's equivalent short circuit but your circuit diagram is not completely correct. Assuming an ideal OpAmp, the corrected schematic including the sneaky \$I_T\$ is as follows







simulate this circuit &ndash; Schematic created using CircuitLab

Now the goal is to find $$\dfrac{V_T}{I_T}$$
  1. \$V_T = -A\cdot V_x - V_x\$ (difference between output and negative input terminal of the OpAmp)
  Thus $$V_T = -V_x(1 + A)$$
  2.\$V_x = -I_T\cdot R_s\$ (\$I_T\$ is an ideal current source so it forces the current in the whole circuit to be \$I_T\$)  

Plugging 2 into 1
  $$V_T = I_T\cdot R_s(1 + A)$$
Dividing both sides by \$I_T\$ yields:
$$\dfrac{V_T}{I_T} = R_T = R_s(1 + A)$$  

You can obtain the same results by placing a Voltage source \$V_T\$ in series with the element and finding the current \$I_T\$.