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:

1+15

1+20

1+12

1+12+12

1+12+15

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.