You can't be sure at this point in your proof, depending on the starting premise, that any other partition $Q$ with mesh $||Q||<\delta$ is necessarily in the set $S$.
What you are trying to prove is usually taken as the definition that $f$ is Riemann integrable.
The function $f:[a,b] \rightarrow \mathbf{R}$ is Riemann integrable with integral value $I$ if for every $\epsilon > 0$ there is a $\delta >0$ such that for any partition $P = (x_0,x_1,\ldots,x_n)$ with $||P|| = \max_{1 \leq i\leq n}(|x_i-x_{i-1}|)< \delta $ and any set of tags $\xi_i \in [x_{i-1},x_i]$ then
$$\left|\sum_{i=1}^{n}f(\xi_i)(x_i-x_{i-1})- I\right|=|S(P,f)-I| < \epsilon.$$
If $f$ is Riemann integrable, then it must be bounded.
So I assume you are trying prove this starting from Darboux's criterion for integrability.
The bounded function $f:[a,b] \rightarrow \mathbb{R}$ is Riemann integrable if the upper and lower Darboux integrals are equal. Or, equivalently, if for any $\epsilon > 0$ there exists a partition $P$ such that $U(P,f)-L(P,f) < \epsilon$
Using your notation:
$$\int_{a}^{b}f=I=U(f) = L(f), $$
where
$$U(f) = \inf_{P} \,U(P,f),\\ L(f)= \sup_{P} \,L(P,f)$$
This implies that for any $\epsilon > 0$ there are partitions $P_1$ and $P_2$ such that $I-\epsilon/2 < L(P_1,f)$ and $U(P_2,f) < I+\epsilon/2.$ Let $P_3 = P_1 \cup P_2$ be a common refinement. Note that a partition $P'$ is a refinement of $P$ if every point in $P$ is also in $P'$. If $P'$ refines $P$ then $||P'|| \leq ||P||$, but the converse is not necessarily true.
Then
$$I-\epsilon/2 < L(P_1,f) \leq L(P_3,f) \leq U(P_3,f) \leq U(P_2,f)< I + \epsilon/2.$$
At this point, you can show that any tagged Riemann sum corresponding to a partition that refines $P_3$ is within $\epsilon$ of $I$, but you still need to show that this is true if the mesh of the partition is sufficiently small regardless of whether or not it refines $P_3.$
Let $D=\sup\{|f(x)−f(y)|:x,y∈[a,b]\}$ denote the maximum oscillation of $f$ and let $δ=ϵ/2mD\,$ where $m$ is the number of points in the partition $P_3$ .
Now let $P$ be any partition with $||P|| < \delta$ . Form the common refinement $Q=P∪P_3$ .
You will see that the upper sums $U(P,f)$ and $U(Q,f)$ differ in at most $m$ sub-intervals and at each the deviation is bounded by $δD$, and
$$|U(P,f)-U(Q,f)| < m\delta D=mD\frac{\epsilon}{2mD}=\epsilon/2$$
It follows that
$$U(P,f)<U(Q,f)+ϵ/2\leq U(P_3,f)+ϵ/2<I+ϵ.$$
By a similar argument, you can show $L(P,f)>I−ϵ$.
Hence for any tagged Riemann sum, $S(P,f)$
$$ I-\epsilon < L(P,f)\leq S(P,f)\leq U(P,f) < I+ϵ,$$
and
$$|I - S(P,f)| < \epsilon.$$
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
For any $ r \in [a,b)$, $f$ is Riemann integrable, and for any $\epsilon > 0$ there exists a partition $P_r$ of $[a,r]$ such that by the Riemann criterion, the difference between upper and lower sums satisfies
$$U(P_r,f) - L(P_r,f) < \epsilon/2.$$
Since $f$ is bounded, we can choose $r$ such that $b-r < \epsilon/(4M)$ and
$$[\sup_{x \in [r,b]}f(x) - \inf _{x \in [r,b]}f(x)](b-r) < 2M\frac{\epsilon}{4M} = \frac{\epsilon}{2} ,$$
where $M = \sup_{x \in [a,b]}|f(x)|$.
Extending the partition $P_r$ to a partition $P$ of $[a,b]$ by adding the point $b$ we have
$$U(P,f) - L(P,f) = U(P_r,f) - L(P_r,f) + [\sup_{x \in [r,b]}f(x) - \inf _{x \in [r,b]}f(x)](b-r)< \epsilon.$$
Thus $f$ is Riemann integrable on $[a,b]$.