That the first condition implies the first is immediate, since (using your notation) you always have $m_i \le f(\xi_i) \le M_i$, so the sums in the second definition are caught between the $L$ und $U$ sums.
Edit in response to a comment an additional explanation is necessary here. For this direction
it suffices to show that $I^* = lim_{||P||\rightarrow 0} L(f,P)$ and $I_* = lim_{||P||\rightarrow 0} U(f,P)$ Since both parts are similar it suffices to show, e.g., the first equality.
First it is easy to see that for partitions $P\subset P^\prime$ we have $L(f,P)\le L(f,P^\prime)$. A remaining hurdle is that for two partitions we do not necessarily know that one is a subset of the other one. This is resolved by looking at common refinements:
Assume $P$ satisfies $L(f,P) > I^* - \varepsilon$ and $Q$ is an arbitrary partition. We need to show that then there is a refinement $Q^\prime$ of $Q$ such that $L(f,Q^\prime)\ge L(f,P)$ (and, consequently, $L(f,Q^\prime)>I^*-\varepsilon$).
For $Q^\prime$ one can choose the common refinement $R$: if $P=\{x_1,\ldots x_n \}$ and $Q=\{y_1,\ldots y_m \}$ then we just let $R = P\cup Q$. Since this is a refinement of both $P$ and $Q$ we have both $L(f,R)\ge L(f,P)$ as well as $L(f,R)\ge L(f,Q)$
Second edit: the original version was not correct:
For the other direction it suffices to show that if the function is integrable in the sense of the second definition then both $I_*$ and $I^*$ agree with the of the sums from the second definition. Since the reasoning is the same in both cases I'll just look at $I_*$.
So fix $\varepsilon >0$ and a given partition $P$ such that
$$|L - \sum_{i=1}^n f(\xi_i)\Delta x_i |< \varepsilon$$
if only the partition is fine enough.
Choose such a partition $P=\{x_0,\dots x_n\}$ and to $[x_{i-1},x_{i}]$ choose $\eta_i\in[x_{i-i},x_{i}]$ such that for
$m_i:=\inf \{ f(x):x\in [x_{i-1},x_i]\} $
we have $$0\le f(\eta_i)-m_i\le \frac{\varepsilon}{2n}$$
Then
\begin{eqnarray}
| L -\sum_{i=1}^n m_i \Delta x_i|
& = & |L- \sum_{i=1}^n f(\eta_i)\Delta x_i + \sum_{i=1}^n f(\eta_i)\Delta x_i
-\sum_{i=1}^n m_i\Delta x_i| \\
&\le & |L- \sum_{i=1}^n f(\eta_i)\Delta x_i| + \sum_{i=1}^n | f(\eta_i)
- m_i|\Delta x_i \\
& < & \frac{\varepsilon}{2} + \sum_{i=1}^n \frac{\varepsilon}{2n}=\varepsilon
\end{eqnarray}
If you 'see' that $0 <L -I_*< L -\sum_{i}m_i \Delta x_i$ then you are done here, otherwise it follows easily from the last estimate that the $\sum_i m_i \Delta x_i$ are, for any partition which is fine enough, $\varepsilon $ close to the fixed real number $L$, which of course implies that the $\sup$ over these sums exists and equals $L$ (here you need to use again the fact that you will approach the $\sup$, if it exists, if the width of the partitions goes to $0$).
In this context "Cauchy integral" has the meaning you know.
It is a fact that if a function is bounded and Cauchy integrable over $[a,b]$, then it is also Riemann integrable over that interval.
It seems that there is no elementary proof of this theorem.
The proof in Kristensen, Poulsen, Reich A characterization of Riemann-Integrability, The American Mathematical Monthly, vol.69, No.6, pp. 498-505, (theorem 1), could be considered elementary because plays only with Riemann sums but is an indigestible game.
Note that there exist unbounded functions Cauchy integrable.
Also the use of regular partitions is enough to define Riemann integral.
See Jingcheng Tong Partitions of the interval in the definition of Riemann integral, Int. Journal of Math. Educ. in Sc. and Tech. 32 (2001), 788-793 (theorem 2).
I repeat that the use of only left (or right) Riemann sums with only regular partitions doesn't work.
Best Answer
This is lengthy, but does eventually attempt to get at the question, albeit indirectly.
When I first took calculus, the business of taking "limits" of Riemann sums bothered me for reasons I couldn't articulate. Many years later, with more study of analysis, the reasons became clear: Limits in calculus are defined for (i) functions of one variable or (ii) real sequences. By contrast, a Riemann sum depends on a partition $P = (x_{i})_{i=0}^{n}$ of an interval and a choice of sample points $(t_{i})_{i=1}^{n}$ with $x_{i-1} \leq t_{i} \leq x_{i}$ for $1 \leq i \leq n$.
In short, an individual Riemann sum depends on an arbitrarily large finite number of variables, and taking a limit always entails letting the number of variables grow without bound (because the mesh of the partition must approach $0$).
Darboux integration of a bounded function $f$ therefore came as a huge conceptual relief: We pick a partition $P = (x_{i})_{i=0}^{n}$ of $[a, b]$, but instead of sampling the integrand $f$, we take lower and upper bounds $$ m_{i} = \inf\{f(t) : x_{i-1} \leq t \leq x_{i}\},\qquad M_{i} = \sup\{f(t) : x_{i-1} \leq t \leq x_{i}\}, $$ and form lower and upper sums $$ L(f, P) = \sum_{i=1}^{n} m_{i}\, \Delta x_{i},\qquad U(f, P) = \sum_{i=1}^{n} M_{i}\, \Delta x_{i}. $$ Now there's no need for "limits"! Instead we take the supremum of the set of lower sums (over all partitions $P$) and the infimum of the set of upper sums, and we declare $f$ to be integrable on $[a, b]$ if these two numbers are equal.
This would be a lengthy and pointless digression in the context of this question but for an elementary observation: For every partition $P = (x_{i})_{i=0}^{n}$, for every subinterval $[x_{i-1}, x_{i}]$, and for every sample point $t_{i}$, we have $$ m_{i} \leq f(t_{i}) \leq M_{i}. $$ Multiplying across by $\Delta x_{i}$ and summing over $i$ shows that every Riemann sum associated to $P$ is bounded below by $L(f, P)$ and bounded above by $U(f, P)$. Particularly, if $f$ is Darboux integrable on $[a, b]$, then $f$ is Riemann integrable on $[a, b]$. The converse is also true: Loosely, by sampling suitably we can make $f(t_{i})$ as close to the infimum $m_{i}$ (or to the supremum $M_{i}$) as we like. (Usually, we like to take small to mean $< \frac{\varepsilon}{4(b-a)}$ or some such.)
What Darboux integration gives us that Riemann integration doesn't (generally) is workable lower and upper bounds for the quantity we wish to define. The moral of this story (if you ask me) is to frame questions about Riemann integrability in terms of Darboux integrability.
One final coda: In condition 2., there's an implicit existential quantifier on $L$, and presumably "Every choice of $t_{i}$" means "in the corresponding subinterval". Even with those adjustments, however, there's the vexation of the term limit, for the reasons noted above. And, there's a technical hitch in the "for every sequence in the set...", because the stated set of all Riemann sums depends only on the integrand $f$ and the interval $[a, b]$, but in order to "probe" integrability we have to restrict ourselves to sequences sampled from partitions whose mesh converges to $0$.