Why is the concentration of protons equal to the concentration of a strong acid?
The statement $[\ce{H^+}] = [\ce{A^-}]$ is an approximation for not too diluted acids. However you can pretty fast tell that this won't hold if you add a strong acid to water so that the acid has a concentration of let's say $\pu{10^-8 mol//l}$. Via the formula $\mathrm{pH} = \log\left([\ce{H^+}]\right)$ this would result in $\mathrm{pH} = 8$ so if that was true, you just made the solution less acidic by adding an acid.  

In fact the total concentration $[\ce{H^+}]$ is given by $[\ce{H^+}] = [\ce{A^-}] + \pu{10^-7 mol//l}$ where the last term stems from the autoionization of water. However in most use-cases $[\ce{A^-}]$ is far greater than $10^{-7}$. In those cases the autoionization of water can be neglected and thus the approximation $[\ce{H^+}] = [\ce{A^-}]$ gives sufficient results.
However, if you are working with really diluted solutions of acids this approximation is no longer applicable (as demonstrated above) which forces you to actually respect the autoionization of water.  

If you want to be on the safe side you can always respect the autoionization. This will always give you the correct pH (even for not so diluted solutions) but you may neglect it if the acid-concentration is high enough (this of course depends on the accuracy of the results you need).  

EDIT:
As @MaxW pointed out in the comment a general expression would be $[\ce{H+}] = [\ce{A-}] + [\ce{OH-}]$ which for a neutral solution at 25°C (thanks to @Martin for pointing this out) becomes the formula proved above.