Why do oxidation numbers work?
First of all they are not really arbitrary.

But the main point, regarding balancing the equations is that: You can keep track of the total number of electrons per species, which is a well defined value. Like, for one permanganate ($\ce{MnO4-}$) you have 25+4*8+1 electrons, for one $\ce{Mn(II)}$ you have 23 electrons, for one water you have 10 electrons, for one $\ce{H+}$ you have 0 electrons. So for this half-equation, you must have 5 electrons on the left side:

$$\ce{8H+ + MnO4- + 5e- -&gt; Mn(II) + 4H2O}$$

or you can say that, here I assume all H atoms are +, all oxygens are -2, and for consistency, Mn in $\ce{MnO4-}$ is +7. I can dismiss H and O's since all are same, and all change is because of Mn, which goes from +7 to +2 and difference is 5 electrons.

You could do this like that as well: O is -4 everywhere, so H's in $\ce{H2O}$ should be +2 and they were +1 on the left side. They donated 8 electrons. Mn in $\ce{MnO4-}$ should be +15 and it is +2 on the right side, which means it accepted 13 electrons. And the difference is 5 electrons, again.

Bottomline: It is all about consistency. If all elements in each and every species sum up to the charge of this species, whatever you say their formal charges are, it will work -because this just another way of keeping track of total electron count.  So it is better to treat most of them constant like, O is -2, H is +1 almost always.