Derivation of relation between acid and base equilibrium constants
$K_\mathrm{a}$ and $K_\mathrm{b}$ should not have a term for the concentration of water inside.
When you dissolve $x$ moles of $\ce{NH3}$ in water (doesn't matter how much $x$ is), both relations


$$K_\mathrm{a} = \frac{[\ce{NH3}][\ce{H3O+}]}{[\ce{NH4+}]} \tag{1}$$

as well as

$$K_\mathrm{b} = \frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]} \tag{2}$$

must simultaneously hold true. You agree that both of these relations must independently hold no matter what concentration of $\ce{NH3}$ (or $\ce{NH4+}$, or $\ce{H+}$, or $\ce{OH-}$) there is, right? That is why they are called equilibrium constants, after all.

So, in any given single solution with one well-defined value of $[\ce{NH3}]$, one well-defined value of $[\ce{NH4+}]$, ... both equations must simultaneously hold. You can then cancel out the terms as necessary to obtain

$$K_\mathrm{a}K_\mathrm{b} = K_\mathrm{w} \tag{3}$$



If you use two different solutions, with different concentrations of all the relevant species, the relation still holds true. However, the maths will be much harder because for example if you have solution 1 with pH = 1, and solution 2 with pH = 10, then $[\ce{H+}]_\mathrm{sol,1}[\ce{OH-}]_\mathrm{sol,2}$ is no longer equal to $K_\mathrm{w}$. The relation $[\ce{H+}][\ce{OH-}] = K_\mathrm{w}$ is only necessarily true if we are talking about one single solution.