First Solution. Since $f(x) = 0$ has no nonnegative real solutions, $b = f(0) > 0$. If $a \ge 0$, then take $g(x) = f(x), h(x) = 1$. Suppose that $a < 0$. Since $f(x) = 0$ has no nonnegative roots, we must have $a^2 < 4b$. Replacing $f(x)$ by $b^{-1} f(x \sqrt{b})$, we may assume that $b = 1$.

Put $c_1 = -a$ and $f_1(x) = f(x) = x^2 - c_1 x + 1$. Note that $c_1^2 < 4$. Put $f_2(x) = f_1(x)(x^2 + c_1 x + 1) = x^4 - c_2 x^2 + 1$, where $c_2 = c_1^2 - 2$. If $c_2 \le 0$, take $g(x) = f_2(x), h(x) = g(x)/f(x)$ to satisfy the conclusion. If $c_2 > 0$, then put
$$f_3(x) = f_2(x)(x^4 + c_2 x^2 + 1) = (x^8 - c_3 x^4 + 1),$$
so $c_3 = c_2^2 - 2$. Proceeding in this way, suppose that $f_k(x) = (x^{2k} - c_k x^{2k-1} + 1)$, where $c_k = c_{k-1}^2 - 2$, and that $c_k \le 0$, then take $g(x) = f_k(x), h(x) = g(x)/f(x)$ to satisfy the conclusion.

If $c_k > 0$, then put
$$f_{k+1}(x) = f_k(x)(x^{2k} + c_k x^{2k-1} + 1) = x^{2k+1} - c_{k+1} x^{2k} + 1.$$

Suppose that $c_t > 0$ for all positive integers $t$.
Then, for a positive integer $n$, $c_n^2 > 2$, $c_{n-1}^2 > 2+\sqrt{2}$, $c_{n-2}^2 > 2+\sqrt{2+\sqrt{2}}$, $c_{n-3}^2 > 2+\sqrt{2+\sqrt{2+\sqrt{2}}}$, ..., and in general $c_{n-s}^2 > 2+\sqrt{2+\sqrt{2+\sqrt{2}+\dots+\sqrt{2}}}$ (s terms 2 under the radical signs). By the AGM inequality, $2+\sqrt{2} > 2\sqrt{2\sqrt{2}} = 2^{1+\frac{1}{2}+\frac{1}{4}}$, and, in general,
$$2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \dots + \sqrt{2}}}} > 2^{1+\frac{1}{2}+\frac{1}{4}} + \dots + \frac{1}{2^{h+1}},$$
where there are $h$ terms 2 under the radical sign. Hence $c_1^2 > 2^{1+\frac{1}{2}+\frac{1}{4}} + \dots + \frac{1}{2^{m+1}}$, for all positive integers $m$.
Since the sum of the geometric progression
$$1 + \frac{1}{2} + \frac{1}{2^2} + \dots + \frac{1}{2^n} + \dots$$
is 2, while $c_1^2 < 4$, this is a contradiction.
Hence $c_t \le 0$ for some positive integer $t$, and the proof is complete.

Solution 2:

This is based on the following observation, viz., if
$$p(x, \theta) = x^2 - 2 \cos \theta x + 1,$$
then
$$\begin{aligned} p(x, \theta)p(x, \theta + \pi) &= (x^2 - 2 \cos \theta x + 1)(x^2 + 2 \cos \theta x + 1) \\ &= x^4 - 2(2 \cos^2 \theta - 1)x^2 + 1 \\ &= x^4 - 2 \cos 2\theta x^2 + 1 \\ &= p(x^2, 2\theta). \end{aligned}$$

To continue, as in the First Solution, it suffices to consider the case $f(x) = x^2 - c x + 1$, where $0 < c < 2$. Select $\theta \in (0, \pi/2)$, so that $c = 2 \cos \theta$. Then $f(x) = p(x, \theta)$ and so
$$f(x)p(x, \theta + \pi) = p(x^2, 2\theta) = x^4 - 2 \cos 2\theta x^2 + 1.$$

If $\cos 2\theta \le 0$, we stop and take $g(x) = p(x^2, 2\theta)$ and $h(x) = p(x, \theta + \pi)$. Otherwise, we repeat the process and use the fact that
$$p(x^2, 2\theta)p(x^2, 2\theta + \pi) = p(x^4, 4\theta).$$

We stop if $\cos 4\theta \le 0$, and put
$$g(x) = p(x^4, 4\theta), \quad h(x) = p(x, \theta + \pi)p(x^2, 2\theta + \pi).$$
Otherwise we continue. We stop at the $n$th stage where $\cos 2^n\theta \le 0$, and $\cos 2^{n-1}\theta > 0$, at which point we let
$$g(x) = p(x^{2^n}, 2^n\theta), \quad h(x) = p(x, \theta + \pi) \prod_{k=1}^{2^n-1} p(x^{2^k}, 2^k\theta + \pi).$$

It's clear that $g, h$ are polynomials whose coefficients are nonnegative, and that $f = g/h$.