Good question. One reason why one might ponder about this is to recall Lebesgue's original defintion, developing measure theory first, then defining the integral for all bounded measurable functions, then extending to unbounded "summable"
functions. What has this to do with Riemann's definition? Why on earth is every Riemann integrable function a bounded measurable function? And (in the other direction) why on earth is there a bounded measurable function that is not Riemann integrable? If you don't ask such questions you weren't paying attention.
Lebegue doesn't help. His definition is so remote from Riemann's that it takes a long time in his thesis to prove the connection. That remoteness also alienated many mathematicians of his time.
The short historical answer (Maybe not satisfying for you).
Volterra in the 1880s gave an example of a function $F:[0,1]\to\mathbb R$
with $f=F'$ bounded but not Riemann integrable.
Lebesgue, motivated by that example, designed an integral guaranteed to include all Riemann integrable functions but also integrate all bounded derivatives. [Indeed he explicitly claimed that as goal.]
Preliminaries: First let's clarify what we are talking about. Take an interval $[a,b]$ and write the following:
${\cal C}[a,b]$ --- all continuous functions on $[a,b]$
${\cal D}[a,b]$ --- all derivatives on $[a,b]$, i.e., $F'=f$ for all points for some function $F$.
$b{\cal D}[a,b]$ --- all bounded derivatives on $[a,b]$
${\cal R}[a,b]$ --- all Riemann integrable functions on $[a,b]$
${\cal IR}[a,b]$ --- all improperly Riemann integrable functions on $[a,b]$
${\cal L}[a,b]$ --- all Lebesgue integrable functions on $[a,b]$
${\cal IL}[a,b]$ --- all improperly Lebesgue integrable functions on $[a,b]$
If you understand your integration theory (and you should before considering such questions as this) you should know these facts intimately:
${\cal C}[a,b] \subset b{\cal D}[a,b] \subset b{\cal L}[a,b]
\subset {\cal L}[a,b] \subset {\cal IL}[a,b]$ and all inclusions proper.
While $ b{\cal D}[a,b] \subset b{\cal L}[a,b]$ it is the case that
$b{\cal D}[a,b] \not\subset {\cal R}[a,b] $. This is the original motivation Lebesgue gave for his integral.
$ {\cal IR}[a,b] \not \subset {\cal L}[a,b]$
and $ {\cal L}[a,b] \not \subset {\cal IR}[a,b]$ although they are compatible, i.e., if a function is integrable in both senses the integrals agree.
But $ {\cal D}[a,b] \not \subset {\cal L}[a,b]$ and $ {\cal D}[a,b] \not \subset {\cal IL}[a,b]$ which is the reason why integration theory did not stop with Lebesgue. [To elaborate a bit: Lebesgue's integral handles all bounded derivatives but only some unbounded derivatives. He did not like this fact!]
Q. Back to your question: What part of the definition
of ${\cal L}[a,b]$ explains why $b{\cal L}[a,b]\supset {\cal R}[a,b]$ and
$b{\cal L}[a,b] \not= {\cal R}[a,b]$?
A1. Well this depends on which of the very many definitions of the Lebesgue integral you are using. If you use the measure-theoretic definition for
${\cal L}[a,b]$ and the orginal definition for ${\cal R}[a,b]$ this is a puzzle. But you are correct: if you use instead the Peano-Jordan measure definition of ${\cal R}[a,b]$ (as you suggested) then the explanation is quite clear. Just check that all PJ-measurable sets are measurable and give one simple example of a measurable set that is not PJ-measurable. Your thinking here is fine!
A2. But you might have seen a different definition for ${\cal L}[a,b]$ as
the completion of the space ${\cal C}[a,b]$ using the norm
$\|f\| = \int_a^b |f(x)|\,dx$. In that case your explanation will need to show that this completion contains ${\cal R}[a,b]$ but that there are bounded functions in that completion not contained in ${\cal R}[a,b]$.
A3. Perhaps you are a fan of defining ${\cal L}[a,b]$ using Riemann sums monitored by a gauge (as developed in the 1960s). Then it is immediate and trivial that ${\cal L}[a,b]\supset{\cal R}[a,b]$ and, again, one bounded counterexample will suffice.
A4. I doubt you want any more definitions for ${\cal L}[a,b]$. So bye.