If a positive function $u:[0,1]\to \mathbb{R}^+$ is known to be continuous and twice differentiable, and to have no local maxima in $(0,1)$ (i.e. $u$ is maximised on the boundary $\{0,1\}$), is it possible to find bounded $g(x)$ and negative $h(x)$ such that the one-dimensional maximum principle applies
$$u''+g(x)u'+h(x)u\geq 0$$
and if so, is there a way to construct such $g,h$?
No, in general this is not possible. If $u$ has a critical point $x_0$ with $u'(x_0) = u''(x_0) = 0$, then the left-hand side at $x_0$ is $h(x_0)u(x_0)<0$. As an example, take $u(x) = 1 + (x-1/2)^3$ and $x_0 = 1/2$.
However, if all critical points are non-degenerate, then your assumption implies that $u''(x_0)>0$ for every $x_0$ with $u'(x_0) = 0$. In this case you can take $g(x) = C u'(x)$ for a large constant $C>0$, and $h(x) = -\epsilon$ for a small constant $\epsilon > 0$ and check that the condition is satisfied.