I try to prove this without invoking anything about Lebesgue integral. Moreover, I do not assume the fact that $fg$ is Riemann integrable nor the inequality $| \int_a^b f(x)g(x) dx | \leq \int_a^b |f(x)g(x)|dx$.
$f$ is Riemann integrable $\Rightarrow$ $f$ is bounded. Choose
$M>0$ such that $|f(x)|\leq M$ for all $x\in[a,b]$. Let $\varepsilon>0$
be given. Choose $\delta>0$ such that for any partition $\mathbb{P}=\{x_{0},x_{1},\ldots,x_{n}\}$
of $[a,b]$ (with $a=x_{0}<x_{1}<\ldots<x_{n}=b$) and any $\xi_{i}\in[x_{i-1},x_{i}]$,
if $||\mathbb{P}||<\delta$ (here, $||\mathbb{P}||=\max_{1\leq i\leq n}|x_{i}-x_{i-1}|$),
then
$$
\left|\sum_{i=1}^{n}g(\xi_{i})(x_{i}-x_{i-1})-\int_{a}^{b}g(x)dx\right|<\frac{\varepsilon}{M}.
$$
That is,
$$
\left|\sum_{i=1}^{n}g(\xi_{i})(x_{i}-x_{i-1})\right|<\frac{\varepsilon}{M}.
$$
Now, let $\mathbb{P}=\{x_{0},x_{1},\ldots,x_{n}\}$ be an arbitrary
partition of $[a,b]$, with $a=x_{0}<x_{1}<\ldots<x_{n}=b$, that
satisfies $||\mathbb{P}||<\delta$. Let $\xi_{i}\in[x_{i-1},x_{i}]$
be arbitrary. Then
\begin{eqnarray*}
& & \left|\sum_{i=1}^{n}f(\xi_{i})g(\xi_{i})(x_{i}-x_{i-1})-0\right|\\
& \leq & \sum_{i=1}^{n}|f(\xi_{i})g(\xi_{i})|(x_{i}-x_{i-1})\\
& \leq & M\sum_{i=1}^{n}g(\xi_{i})(x_{i}-x_{i-1})\\
& < & M\cdot\frac{\varepsilon}{M}\\
& = & \varepsilon.
\end{eqnarray*}
This shows that the Riemann integral $\int_{a}^{b}f(x)g(x)dx$ exists
and $\int_{a}^{b}f(x)g(x)dx=0$.