Answer by periblepsis for Proving the output voltage expression

incorrect diagram & all results incorrect

First, let me correct your picture.

The diagram below on the left is what you are showing and it won't fit your assumption that \$V_p=V_m\$. The reason is that the output responds in a positive direction to a positive difference (your \$V_d\$) and if the output moves more positively then the difference, \$V_d\$, would become still more positive.

The positive feedback loop here would not result in \$V_p=V_m\$.

^{simulate this circuit– Schematic created using CircuitLab}

The diagram above on the right does (in the ideal -- no offset voltage) meet the assumption that \$V_p=V_m\$ as now there is negative feedback being applied.

So, despite the fact that I think you followed through with valid KCL logic, you started with a false assumption. So you reached a wrong result for that reason.

But close.

And I don't think the highlighted result is correct, either.

provide sources

Which gets to an issue. When writing you questions out like this, please provide source information. Even if it isn't available via a web link it's still possible that someone may have the text and can confirm you accurately expressed it.

KCL

Here's my KCL for the right-side schematic (the correct one, not the incorrect one which only wastes time worrying over.) I usually want to apply KCL to the output pin, as well, which is why I included \$I_s\$ in my right-hand schematic. Keeps things clean (to me):

$$\begin{align*}\frac{V_m}{R_1}+\frac{V_m}{R_2}+I_m&=\frac{V_s}{R_2}\\\\\frac{V_p}{R_3}+I_p&=0\:\text{A}\\\\\frac{V_s}{R_2}&=I_s+\frac{V_m}{R_2}\end{align*}$$

Now, that's three equations and four unknowns. But \$V_p=V_m\$ is the fourth equation (ideal assumption.) So we have enough.

From that, I find that:

$$V_s=I_m\cdot R_2-I_p\cdot R_3\cdot\left(1+\frac{R_2}{R_1}\right)$$

This is quite close to your answer. It's just that you used a bad diagram and so assigned the wrong bias currents. But otherwise, I think you were able to move in the right direction.

added note

Since the closed loop voltage gain for the correctly wired circuit is \$A_v=-\left(1+\frac{R_2}{R_1}\right)\$, the above equation can be re-phrased with "lower entropy" (as Basso writes it):

$$V_s=I_m\cdot R_2+I_p\cdot A_v\cdot R_3$$

For this case, it suggests that the closed loop voltage gain arranged at the (-) input magnifies and inverts the impact of \$I_p\$ on \$R_3\$ at the (+) input.