You can generalize your "weaker" definition as follows.
Let $\mathcal{I}$ denote the set of closed intervals $[a,b]$ and let $\tau : \mathcal{I} \to \mathbb{R}$ be any function such that $\tau([a,b]) \in [a,b]$.
Examples are $\tau([a,b]) = a$, $\tau([a,b]) = b$, $\tau([a,b]) = \frac{a + b}{2}$ etc.
Define $S(f,P,\tau) = \sum_{i=1}^n f(\tau([x_{i-1},x_i])) (x_i-x_{i-1})$. Let us say that $f$ is $\tau$-integrable with $\tau$-integral $K$ if the obvious condition is satisfied.
As pointed out by Paramanand Singh, the following are equivalent for a bounded function $f$:
(1) $f$ is Darboux integrable
(2) $f$ is Riemann integrable
(3) $f$ is $\tau$-integrable for all $\tau$
(4) $f$ is $\tau$-integrable for some $\tau$
$\tau$-integrability seems to be conceptually simpler than Riemann integrability because it avoids to use tags. But it should be clear that there is a substantial arbitrariness in the choice of $\tau$. It can be defined by a simple rule as in your question, but it can also be "erratic". It may even depend on $f$ if you want. For example, if $f$ is continuous, then you can take $\tau([x_{i-1},x_i]))$ to be any point of $[x_{i-1},x_i]$ at which $f \mid_{[x_{i-1},x_i]}$ attains its minimum (or maximum).
So what might be the benefit of general tags? Here are some arguments.
a) If you know that $f$ is integrable (for example if $f$ is continuous or monotonic), then you can choose suitable tags $T$ which make $S(f,P,T)$ explicitly evaluable. For example, you can prove that $\int_0^t \frac{1}{1 + x^2}dx = \arctan t$ by choosing a tag as $\xi_i \in [x_{i-1},x_i]$ such that $\frac{1}{1 + \xi_i^2} = \frac{1}{1 + x_{i-1}x_i}$. I shall not go into details and I do not claim that this is an elegant proof, but it shows that general tags can be useful.
b) If you have integrable functions $f,g$, you know that their product $fg$ is integrable. Then you may choose any two $\xi_i, \xi'_i \in [x_{i-1},x_i]$ and consider the sums
$$\Sigma_{i=1}^n f(\xi_i)g(\xi'_i)(x_i - x_{i-1}) .$$
These converge to $\int_a^b f(x)g(x)dx$ as was shown by G.A. Bliss.
The idea is pretty clear, but we just have to write a lot, and use a lot of notation.
Definitions. A partition of $[a,b]$ is a finite sequence $\pi = (a_j)_{j=0}^J$ with
$a=a_0 < a_1 < \cdots < a_J = b$.
The norm of $\pi$ is $\|\pi\| = \max_{1 \le j \le J} |a_j-a_{j-1}|$.
A tagged partition of $[a,b]$ is a pair $(\pi,\gamma)$ where $\pi = (a_j)_{j=0}^J$
is a partition of $[a,b]$ and the finite squence $\gamma = (c_j)_{j=1}^J$ satisfies
$a_{j-1} \le c_j \le a_j$ for all $j$.
Let $f : [a,b] \to \mathbb R$. Let $\pi = (a_j)_{j=0}^J$ be a partition of $[a,b]$.
The upper sum is
$$
U(f,\pi) = \sum_{j=1}^J M_j(f)\;\cdot(a_j-a_{j-1})\qquad\text{where}\qquad
M_j(f) = \sup \{f(x) : a_{j-1} \le x \le a_j\} .
$$
The lower sum is
$$
L(f,\pi) = \sum_{j=1}^J m_j(f)\;\cdot(a_j-a_{j-1})\qquad\text{where}\qquad
m_j(f) = \inf \{f(x) : a_{j-1} \le x \le a_j\} .
$$
Let $(\pi,\gamma)$ be a tagged partition. The Riemann sum is
$$
R(f,\pi,\gamma) = \sum_{j=1}^J f(c_j)\cdot (a_j-a_{j-1})
$$
We say that $f$ is Riemann integrable, and its integral is $V$ iff
for every $\epsilon > 0$ there exists $\delta > 0$ such that for every
tagged partition $(\pi,\gamma)$ of $[a,b]$, if $\|\pi\| < \delta$, then
$$
\big |R(f,\pi,\gamma) - V \big| < \epsilon .
$$
Alternatively (from a Cauchy criterion), $f$ is Riemann integrable if and only if:
for every $\epsilon > 0$ there exists $\delta > 0$ such that for every
partition $\pi$ of $[a,b]$, if $\|\pi\| < \delta$ then
$$
U(f,\pi) - L(f,\pi) < \epsilon .
$$
If so, then the integral $V$ is the unique number with
$L(f,\pi) \le V \le U(f,\pi)$ for all partitions $\pi$ of $[a,b]$.
Theorem A
Let $f_1,f_1,\dots, f_K$ be functions on $[a,b]$ such that $f_1, f_2,\dots, f_K$
and the product $F = f_1 f_2\cdots f_K$
are all Riemann integrable.
Claim: for any $\epsilon > 0$ there exists $\delta > 0$ such that: if $\pi$ is a partition of $[a,b]$ and $\|\pi\| < \delta$ and $\gamma^{(k)}=(c_j^{(k)})_{j=1}^J$ , $1 \le k \le K$, are $K$ choices of tags so that for all $k$, the pair $(\pi,\gamma^{(k)})$ is a tagged partition of $[a,b]$, then
$$
\left|U(F,\pi) - \sum_{j=1}^J f_1(c_j^{(1)}) f_2(c_j^{(2)})\cdots
f_K(c_j^{(K)})\cdot (a_j-a_{j-1})\right| < \epsilon
\tag1$$
and
$$
\left|\sum_{j=1}^J f_1(c_j^{(1)}) f_2(c_j^{(2)})\cdots
f_K(c_j^{(K)})\cdot (a_j-a_{j-1})-L(F,\pi)\right| < \epsilon
\tag2$$
and
$$
U(F,\pi) - L(F,\pi) < \epsilon.
\tag3$$
Case 1: All $f_k \ge 0$.
So, $\epsilon > 0$ is given. Since all the functions $f_k$ are Riemann integrable, they are bounded. There is a single bound $A > 0$ with
$$
|f_k(x)|\le A \qquad\text{for}\qquad k=1,\dots,K,\qquad a\le x \le b .
$$
Let $$\epsilon' = \frac{\epsilon}{K A^{K-1}}.$$
Since $f_1,\dots,f_K$ are all Riemann integrable, there is $\delta > 0$ so that
for any partition $\pi = (a_j)_{j=0}^J$ of $[a,b]$, if $\|\pi\| < \delta$, then
$$
U(f_k,\pi) - L(f_k,\pi) < \epsilon'\qquad\text{for } k=1,\dots,K .
$$
Let $(\pi, \gamma^{(k)})$ be tagged partitions as above. Assume $\|\pi\| < \delta$. Write $\widetilde{M}_j = M_j(f_1)M_j(f_2)\cdots M_j(f_K)$ and
$\widetilde{m}_j = m_j(f_1)m_j(f_2)\cdots m_j(f_K)$.
Write
$$
\widetilde{U} = \sum_{j=1}^J \widetilde{M}_j\cdot(a_j-a_{j-1}),\qquad
\widetilde{L} = \sum_{j=1}^J \widetilde{m}_j\cdot(a_j-a_{j-1})
$$
Claim 1: $\widetilde{U}-\widetilde{L} < \epsilon$
Claim 2: All three of
$$
U(F,\pi), \quad L(F,\pi), \quad
\sum_{j=1}^J f_1(c_j^{(1)}) f_2(c_j^{(2)})\cdots
f_J(c_j^{(K)})\cdot (a_j-a_{j-1})
$$
are between $\widetilde{U}$ and $\widetilde{L}$.
Consequently, we obtain $(1), (2), (3)$, as required.
Proof of Claim 1.
The difference $\widetilde{U}-\widetilde{L}$ is written as a sum of $L$ terms. First, fix $j$ between $1$ and $J$. Then
\begin{align}
\widetilde{M}_j &= M_j(f_1)M_j(f_2)M_j(f_3)\cdots M_j(f_K)
\\ & \ge m_j(f_1)M_j(f_2)M_j(f_3)\cdots M_j(f_K)
\\ & \ge m_j(f_1)m_j(f_2)M_j(f_3)\cdots M_j(f_K)
\\ & \ge \cdots
\\ & \ge m_j(f_1)m_j(f_2)\cdots m_n(f_K) = \widetilde{m}_j
\end{align}
From each row to the next, one $M_j$ changes to an $m_j$. Then
\begin{align}
\widetilde{M}_j - \widetilde{m}_j &=
(M_j(f_1)-m_j(f_1)) M_j(f_2)M_j(f_3)\cdots M_j(f_K)
\\ &+ m_j(f_1)(M_n(f_2)-m_j(f_2)))M_j(f_3)\cdots M_j(f_K)
\\ &+ \cdots
\\&+ m_j(f_1) m_j(f_2)\cdots m_j(f_{K-1})(M_j(f_K)-m_j(f_K))
\\ &\le A^{K-1}\sum_{k=1}^K (M_j(f_k) - m_j(f_k))
\end{align}
Now multiply by $a_j-a_{j-1}$ and sum on $j$. The left side is
$$
\sum_{j=1}^J (\widetilde{M}_j - \widetilde{m}_j)\cdot(a_j-a_{j-1})
=\widetilde{U} - \widetilde{L}.
$$
The right side is
\begin{align}
A^{K-1}\sum_{k=1}^K &\sum_{j=1}^J (M_j(f_k) - m_j(f_k))\cdot(a_j-a_{j-1})
= A^{K-1}\sum_{k=1}^K \big(U(f_k,\pi) - L(f_k,\pi)\big)
\\ & < A^{K-1} K \epsilon' =\epsilon
\end{align}
This completes the proof of Claim 1.
Proof of Claim 2.
To prove $\widetilde{U} \ge U(F,\pi) \ge \widetilde{L}$ note that
$\widetilde{M}_j \ge M_j(F) \ge \widetilde{m}_j$.
To prove $\widetilde{U} \ge L(F,\pi) \ge \widetilde{L}$ note that
$\widetilde{M}_j \ge m_j(F) \ge \widetilde{m}_j$.
To prove
$$
\widetilde{U} \ge \sum_{j=1}^J f_1(c_j^{(1)}) f_2(c_j^{(2)})\cdots
f_J(c_j^{(K)})\cdot (a_j-a_{j-1}) \ge \widetilde{L}
$$
note
\begin{align}
\widetilde{M}_j &= M_j(f_1)M_j(f_2)\cdots M_j(f_K)
\\ &\ge f_1(c_j^{(1)}) f_2(c_j^{(2)})\cdots f_K(c_j^{(K)})
\\ &\ge m_j(f_1)m_j(f_2)\cdots m_j(f_K) = \widetilde{m}_j
\end{align}
multiply by $a_j-a_{j-1}$ and sum on $j$.
Case 2. arbitrary signs.
Write $f_k = f_k^+ - f_k^-$ in positive and negative parts. Then
$f_1 f_2\cdots f_k$ is a linear combination of $2^K$ such sums with nonnegative functions. Both
$$
\sum_{j=1}^J f_1(c_j^{(1)}) f_2(c_j^{(2)})\cdots
f_K(c_j^{(K)})\cdot (a_j-a_{j-1}) \qquad\text{and}\qquad
\int_a^b f_1(x)f_2(x)\cdots f_K(x)\;dx
$$
are linear combinations of $2^K$ terms of the same form, but involving only nonnegative functions. Theorem A for $f_1f_2\cdots f_k$ follows from those $2^k$ cases of Theorem A for nonnegative functions.
Best Answer
By a fixed tag, I'm assuming you mean that for each interval of a partition, you get to pick a specific point of your choice.
If a function satisfies your fixed tag definition then it may still not be Riemann integrable. The function $f : [0, 1]\to [0, 1]$ taking value 1 on rationals and 0 on irrationals is not Riemann integrable. However, it will satisfy your definition and have integral of 1 if your fixed tag is always rational and integral of 0 if your fixed tag is always irrational.