We can express the inverse zeta function $$ \prod_{i} (1 - x_i^{-s}) = \sum_ x \mu(x) x^{-s} = \int_1^{\infty} \left(\sum_{y \leq x } \mu(y) \right) s x^{-s-1} dx $$

From this integral representation, we immediately get the implication that the RH in the form $\sum_{y \leq x } \mu(y) = O( x^{ 1/2+ \delta})$ for all $\delta>0$ implies RH in the form that the inverse zeta function is holomorphic for $\operatorname{Re}(s) > 1/2$, as then the integral representation is convergent in that range. I believe the reverse implication can be proven by the usual contour integral argument, maybe with some assumption that the $x_n$ do not grow too much more slowly than the primes, so that the zeta function can be controlled to the right of the critical strip.

Similary, the prime number theorem in the form $\sum_{y \leq x}\mu(y) = o(x)$ implies the inverse zeta function is $ \int_{1}^{\infty} s o (x^{-s} ) dx = o ( |s|/ (\operatorname{Re} s -1))$ where the $o$ goes to $0$ as $\operatorname{Re}(s)$ goes to $1$. This is because $ \int_{1}^{\infty} s x^{-s} dx = s / (s-1)$ and as $\operatorname{Re} s$ goes to $1$, most of the mass is supported on large values of $x$ where $o(x)$ is much smaller than $x$. I'm not sure if this necessary condition for PNT is also sufficient or if more is needed.

For questions 4, 5, 6, and 7, the game is going to be to work with the dlog of the zeta function

$$ \sum_i \frac{ d \log (1- x_i^{-s})^{-1} }{ds} = \sum_i \sum_{j=1}^{\infty} x_i^{-js} \log x_i = \sum_{j=1}^{\infty} \int_1^{\infty} \left( \sum_{x_i \leq x^{1/j}} \log x_i \right) s x^{-s-1} dx $$

and plug in estimates for the sum to evaluate the integral.

For question (4), the sum $ \sum_{y \leq x^{1/j}} \log x_i $ will of course be close to half the number of primes up to $x^{1/j}$, with square-root error. The square-root error will integrate to a holomorphic function in the $\operatorname{Re }s >1/2$ region, so the Beurling zeta function will be equal to the square root of the usual zeta function up to a nonvanishing holomorphic factor. In particular, its inverse will not be holomorphic, as it has nontrivial monodromy around $s=1$. If we instead take two copies of $p_i$ with probability $1/2$ and no copies with probability zero, then the Riemann hypothesis for the modified zeta function follows from Riemann for the original zeta function.

For question (5) and (7), I think something similar happens. We have a precise asymptotic for the number of Beurling primes up to $x$, with small error term, but the main term looks different from the main term in the prime number theorem, and so we won't get a Riemann hypothesis the way you've formulated it.

For example, for question (7), if we take $x_n$ to be a small perturbation of $n$, then the log of the inverse zeta function is close to $\sum_n \log (1- n^{-s}) = -\sum_n \sum_{j=1}^{\infty} n^{-js} / j = -\sum_{j=1}^{\infty} \zeta(js) / j$. For $s$ near $1$, the dominant term is $-\zeta(s)$, which is close to $-1/(s-1)$, so the inverse zeta function itself is equal to $e^{-1/(s-1)}$ up to a nonvanishing holomorphic factor. When the real part of $s$ is positive, the real part of $-1/(s-1)$ is always negative, so the growth rate of this function is not so large and I'm not sure whether (1) holds, but as there is an essential singularity at (2),

However, it's possible to give a sequence with a non-arithmetic definition that unconditionally satisfies the Riemann hypothesis.

If we define $x_n$ inductively by $x_1 =2$, $x_{n+1} = x_n + \log x_n$ for $n>1$, then $\sum_{i=1}^n \log x_i = x_{n+1} - 2$ so $\sum_{x_i \leq x} \log x_i = x + O (\log x)$ which means the logarithmic derivative of the zeta function is close to $s/(1-s)$. Exponentiating, the inverse zeta function is going to be $(s-1)$ times something holomorphic on $ \operatorname {Re} s>1/2$, which by a contour integral should give the Riemann hypothesis.

If we take half the primes at random, the zeta function we get will be, up to a nonvanishing holomorphic factor, the square root of the usual zeta function. So zeta inverse will, near $s=1$, look like $(s-1)^{1/2}$. This has a singularity at $s=1$, but a very mild one - in particular the function grows slower than $1/(s-1)$. I suspect when you do the contour integral to estimate the Mobius sum here, you get the prime number theorem, i.e. it satisfies (1).