Why does the ionic product of water remain constant after addition of non-neutral solute?
$K_w$ is the equilibrium constant of this reaction:
$$\ce{2H2 O &lt;=&gt; H3O+ +OH-}.$$ 
So, it's constant at a fixed temperature. It equals $10^{-14}$ at $25~^\circ\mathrm{C}$. At this temperature $\ce{[H3O+]=[OH^{-}]}=10^{-7}~\mathrm{mol/L}$.

If you add acid to water, the above reaction equilibrium is  displaced in order to counteract the change in the concentration of ion hydronium, i.e. the reaction is displaced to the left. This means that  ions hydroxide will react with the excess of ions hydronium to give water. As a result: the concentration of ion hydroxide will be less than $10^{-7}~\mathrm{mol/L}$, and the concentration of ion hydronium will be more than  $ 10^{-7}~\mathrm{mol/L}$. But their product is constant. $\ce{[H3O+] \cdot[OH^{-}]}=10^{-14}$.