If such a G of order 60 has that class equation, it indeed must be simple. The main point of your proof (although it is hard to understand) is this:
A normal subgroup must be a union of conjugacy classes (since a normal subgroup contains conjugates of all its elements). But the following sums are not divisors of 60:
any other sums will total more than 30, and if a subgroup contains more than half the elements of a given group, it must contain the entire group.
thus the only PROPER subgroup that is normal is the trivial subgroup, hence G is simple.