Answer: $c = \pm\frac{1}{2}$.

Note that
$$\begin{align*} f(c) \cdot f(-c) - f(a) &= (c^2 - 2a c + b)(c^2 + 2a c + b) + a^2 - b \\ &= (c^2 + b)^2 - 4a^2 c^2 + a^2 - b. \end{align*}$$
Putting $d = c^2$ we get
$$\begin{align*} f(c) \cdot f(-c) - f(a) &= (d + b)^2 - 4a^2 d + a^2 - b \\ &= a^2(1 - 4d) + \left(b + d - \frac{1}{2}\right)^2 + d - \frac{1}{4}. \end{align*}$$
If $d < \frac{1}{4}$ then taking $a = 0$, $b = -d + \frac{1}{2}$ we get a contradiction:
$f(c) \cdot f(-c) - f(a) = d - \frac{1}{4} < 0$, a contradiction.

If $d > \frac{1}{4}$ then taking $a = 1$, $b = -d + \frac{1}{2}$ we get a contradiction:
$f(c) \cdot f(-c) - f(a) = -3\left(d - \frac{1}{4}\right) < 0$.

$$\text{If } d = \frac{1}{4} \text{ then } f(c) \cdot f(-c) - f(a) = \left(b - \frac{1}{2}\right)^2 \ge 0.$$
We are done.