Some kind of a converse of Leibniz-Newton

If we were lucky enough to have $f$ continuous, we could do this: Define

$$h(x) = G(x)-G(a) -\int_a^x f(t)\,dt.$$

Now $G'=g$ is given. And by the FTC, the derivative of the integral function is $f(x).$ Thus

$$h'(x) =g(x)-f(x)\ge 0.$$

Thus $h$ is nondecreasing on $[a,b].$ But $h(a)=h(b)=0.$ A nondecreasing function that is $0$ at the endpoints must be identically $0.$ Therefore $g\equiv f$ and $g$ is Riemann integrable.