Sort of, and it's a little more complicated than all that. It's easier to compute thrust first, then convert that into prop wash speed.
Typically, prop blades aren't symmetrical, and their step angle varies along the length of the blade. But suppose we take your simple approximation. In that case, you should probably treat each blade as a thin airfoil, and that's relatively easy to solve.
You will need the following pieces of information.
1) Number of blades, N. (2 in your case, but we might as well keep it general.)
2) Pitch angle, θ for each blade.
3) Length, L for each blade.
4) Surface area, A for each blade.
5) Angular velocity, ω for the prop.
6) Forward velocity, v for the prop/aircraft.
7) Air density, ρ for ambient air.
First, you must find angle of attack, α at some point r along the blade. It will vary with velocity of the aircraft, prop velocity, step angle, and position along the blade. The equation governing angle of attack is as follows.
tan(alpha-\theta) = -\frac{v}{r\omega}
Because at high α the prop stalls, we can use approximation for α<<1, tan(α)=α. That way, the above becomes equation for α(r).
Next, we compute lift coefficient using thin airfoil theory. According to such, the lift coefficient CL=2πα. So that's easy enough. Lift generated by a surface element dA of the prop is given by very simple equation.
dF = P_d C_L dA
Here, Pd is dynamic pressure, which is given by another simple formula.
\frac{1}{2}\rho v_{air}^2
The velocity, vair is the vector sum of aircraft velocity and prop speed, so it will also be a function of position r.
v_{air}^2 = v^2 + \omega^2r^2
One final approximation is for the shape of the blades. We will assume they are uniform in width, so dA=dr A/L. With this in mind, and multiplying lift by N to account for number of blades, we can put it all together.
dF = N \frac{1}{2}\rho (v^2 + \omega^2 r^2) 2 \pi \left(\theta-\frac{v}{r\omega}\right) \frac{A}{L} dr
If you try to integrate this directly, you will notice that the integral diverges near r=0. But the reason it does so is because angle of attack there increases past critical, and with approximation used above, lift goes to infinity. That's wrong. Prop there is actually stalled and produces no lift. So the integral must be taken from v/(rω)<<1. So for simplicity, I'll just integrate for r from v/ω to L. With help of Mathematica, we have.
F = \frac{N A \pi \rho \left( \left(3-8\theta+6 ln\left(\frac{v}{\omega L}\right) \right)v^3 +6 L\theta\omega v^2 - 3L^2\omega^2 v + 2L^3\theta\omega^3 \right)}{6L\omega}
It will probably be enough to simplify the above to leading order in v.
F = N A \pi \rho\frac{2L^2\omega^2\theta - 3L\omega v}{6}
You can then get a rough estimate of the wash speed, vw.
\rho v_{w}(v_{w}-v) = F
Keep in mind that this is very rough, but considering all other approximations made, it should be about as good as you can hope for.