Solution:

Let the zeroes be $r$ and $s.$ Using Vieta's formulas, we have that $a = r + s$ and $2a = rs.$

Then we get that $$rs = 2(r + s),$$ which rearranges to $$rs - 2r - 2s = 0$$ $$rs - 2r - 2s + 4 = 4$$ $$(r - 2)(s - 2) = 4.$$

The only possible pairs $(r - 2, s - 2)$ that work are $$(1, 4), (-1, -4), (4, 1), (-4, -1),$$ $$(2, 2), (-2, -2).$$ For any of these pairs, we have that $$a = r - 2 + s - 2 + 4.$$ We want all the such unique values of $a.$ We get that they are $$-1, 0, 8, 9.$$ The sum of these values is $-1 + 8 + 9 = 16.$

Thus, C is the correct answer.