In Munkres's book Calculus on Manifold, page 93, he gives a theorem:
Let $Q$ be a rectangle in $R^n$, let $f: Q\to \mathbb{R}$ be a bounded function. Let $D$ be the set of points of $Q$ at which $f$ fails to be continuous. Then $\int_Q f$ exists (it means Riemann integral exists) if and only if $D$ has (Lebesgue) measure zero in $\mathbb{R}^n$.
But he does not give a similar version theorem when the domain is a Jordan measurable set (a subset $S$ of $\mathbb{R}^n$ is Jordan measurable if and only if $S$ is bounded and boundary of $S$ has Lebesgue measure zero).
He only gives the following theorem below on page 112:
If $S$ is rectifiable (Jordan measurable), and $f: S\to \mathbb{R}$ is a bounded continuous function, then $f$ is integrable over $S$.
But it is not a necessary and sufficient condition (only sufficient).
Question: Do we have the conclusion below?
Let $S \subset \mathbb{R}^n$ be rectifiable (Jordan measurable), $f: S\to \mathbb{R}$ a bounded function, and $D$ the set of points of $S$ at which $f$ fails to be continuous. Then $f$ is integrable over $S$ if and only if $D$ has (Lebesgue) measure zero in $\mathbb{R}^n$.
I am not sure whether this is correct.
The general case is that, when $S$ is also a compact set in $\mathbb{R}^n,$ your guess is correct. But in general, $S$ may not be a compact set, for example, $S=\mathbb{Q}^n\cap [0,1]^n.$ However, the Riemann integrability of $f$ is always equal to the Riemann integrability of $\overline{f},$ here $E$ is a large rectangle satisfying $\overline{S}\subset E,$ and $\overline{f}:E\to \mathbb{R},$ satisfys $ \overline{f}(x)=f(x),\forall\ x\in S, $ and $\overline{f}(x)=0,\forall x\in (E-S).$
Let $f(x)=1,\forall\ x\in S,$ and let $S=\mathbb{Q}^n\cap [0,1]^n,$ then $f$ is continuous on $S,$ but we can see $\overline{f}$ is not continuous at every point of $\partial S.$ But we have $\partial S=[0,1]^n,$ so the Lebesgue measure of $\partial S$ is $1$. So function $\overline{f}$ is not Riemann integrable on $E,$ which means function $f$ is also not Riemann integrable on $S.$