Transfer function of an op amp circuit
Voltage division isn't  a great approach to hang this on. Just build the output from the bottom up.  You just need to know that current doesn't  enter the input terminals of an op amp.

You know Vc, so you know the current through R3.  That has to be the same as the current through  R4, so now you know Vb.  You also know Va, so now you can calculate current through R2, which has to be the same as the current through R1, which would give you Vs.

Looks a little like the input stage of an instrumentation amplifier,but there's an extra resistor.  CORRECTION  This is a two op-amp instrumentation amplifier.  Full discussion of the circuit at http://www.analog.com/static/imported-files/design_handbooks/5812756674312778737Complete_In_Amp.pdf on page 2-4.  

 -- though the resistor numbers and inputs are not quite the same as what you use

Taking a shot at your derivation, starting from the line i2=i1=V2/R3, lets have a go at it

$$ V_B = V_2 + i_2R_4 = V_2 + \frac{V_2R_4}{R_3} $$
$$ = V_2 \left ( 1+ \frac  {R_4}{R_3} \right ) $$

$$ i_3 = \frac{V_A-V_B}{R_2} = \frac{V_1 - V_2 \left ( 1 + \frac{R_4}{R_3} \right )}{R_2}$$

$$V_S= V_1 + i_4 R_1 =V_1 + \frac{V_1R_1}{R_2}  -  V_2 \frac{R_1}{R_2} \left ( 1+ \frac{R_4}{R_3}\right )$$
 or just shifting to make this look a bit more differential:
$$ V_S = V_1 \left ( 1 + \frac{R_1}{R_2} \right ) - V_2 \frac{R_1}{R_2} \left ( 1 + \frac{R_4}{R_3} \right )$$

That's a quick pass, and something feels wrong.  Feel free to correct.