Setting $ y = -x, $ we get \[2f(x^2) = f(0)^2 + 2x^2\]for all $ x $. Setting $ x = 0 $ in this equation, we get $ 2f(0) = f(0)^2, $ so $ f(0) = 0 $ or $ f(0) = 2 $. Suppose $ f(0) = 2 $. Then \[2f(x^2) = 4 + 2x^2,\]so $ f(x^2) = x^2 + 2 $ for all $ x $. In other words, $ f(a) = a + 2 $ for all $ a \ge 0 $. Setting $ x = y = 1 $ in $ f(x^2) + f(y^2) = f(x + y)^2 - 2xy, $ we get \[1^2 + 2 + 1^2 + 2 = (2 + 2)^2 - 2 \cdot 1 \cdot 1,\]which simplifies to $ 6 = 14, $ contradiction. Otherwise, $ f(0) = 0 $. Then $ 2f(x^2) = 2x^2, $ so $ f(x^2) = x^2 $ for all $ x $. In other words, $ f(a) = a $ for all $ a \ge 0 $. Setting $ y = 0 $ in $ f(x^2) + f(y^2) = f(x + y)^2 - 2xy, $ we get \[f(x^2) = f(x)^2.\]But $ f(x^2) = x^2, $ so $ f(x)^2 = x^2 $. Hence, $ f(x) = \pm x $ for all $ x $. Then the given functional equation becomes \[x^2 + y^2 = f(x + y)^2 - 2xy,\]or \[f(x + y)^2 = x^2 + 2xy + y^2 = (x + y)^2.\]We have already derived this, so as far as the given functional equation is concerned, the function $ f(x) $ only has meet the following two requirements: (1) $ f(x) = x $ for all $ x \ge 0, $ and $ f(x) = \pm x $ for all $ x < 0 $. Then we can write \begin{align*} S &= f(0) + (f(1) + f(-1)) + (f(2) + f(-2)) + (f(3) + f(-3)) + \dots + (f(2019) + f(-2019)) \\ &= 2(c_1 + 2c_2 + 3c_3 + \dots + 2019c_{2019}), \end{align*}where $ c_i \in \{0,1\} $. We can check that $ c_1 + 2c_2 + 3c_3 + \dots + 2019c_{2019} $ can take on any value from 0 to $ \frac{2019 \cdot 2020}{2} = 2039190, $ giving us $ \boxed{2039191} $ possible values of $ S $.