You are effectively replacing $\int_a^b f(x) dg(x)$ with $\int_a^b f(x) g'(x) dx$. A quick trip to Wikipedia reveals that this is perfectly fine when $g$ is absolutely continuous, or when it has a continuous derivative, but may not be valid otherwise:
If $g$ should happen to be everywhere differentiable, then the integral may still be different from the Riemann integral
$$\int_a^b f(x) g'(x) \, dx,$$
for example, if the derivative is unbounded. But if the derivative is continuous, they will be the same. This condition is also satisfied if $g$ is the (Lebesgue) integral of its derivative; in this case $g$ is said to be absolutely continuous.
However, $g$ may have jump discontinuities, or may have derivative zero almost everywhere while still being continuous and increasing (for example, $g$ could be the Cantor function), in either of which cases the Riemann–Stieltjes integral is not captured by any expression involving derivatives of $g$.
Edit: I didn't notice the word "smooth" in your question. In this light, the answer to your question is yes.
Also, MathWorld is probably a better reference: the two expressions are equal "if $f$ is continuous and $g'$ is Riemann integrable over the specified interval".