That the existence of a natural (relative) density implies the existence of a (relative) logarithmic density — these two are then of course equal — holds more generally.
For $A \subset \mathbb{N}$ (using the convention $0 \notin \mathbb{N}$ here), define
\begin{align}
N_A(x) &:= \# \{ a \in A : a \leqslant x\}\,,\\
L_A(x) &:= \sum_{\substack{a \in A \\ a \leqslant x}} \frac{1}{a}\,.
\end{align}
Then if we have a substantial $S \subset \mathbb{N}$, that is, a set with $\lim_{x \to \infty} L_S(x) = \infty$, for all $B \subset S$ the inequalities
$$\liminf_{x \to \infty} \frac{N_B(x)}{N_S(x)} \leqslant \liminf_{x \to \infty} \frac{L_B(x)}{L_S(x)} \leqslant \limsup_{x \to \infty} \frac{L_B(x)}{L_S(x)} \leqslant \limsup_{x \to \infty} \frac{N_B(x)}{N_S(x)} \tag{$\ast$}$$
hold. This answers your "aside 1" in the affirmative, and yields the implication stated above, since the left and right terms of $(\ast)$ are equal if $B$ has a (relative) natural density in $S$. We can prove $(\ast)$ via Abel summation:
Let $c$ and $C$ be the left and right hand side of $(\ast)$ respectively. If $c = 0$ there's nothing to prove for the left inequality, otherwise for every $0 < \gamma < c$ there is an $x_{\gamma}$ with $N_B(x) \geqslant d\cdot N_S(x)$ for all $x \geqslant x_{\gamma}$. Then, for $x > x_{\gamma}$ we have
\begin{align}
L_B(x) &= \frac{N_B(x)}{x} + \int_1^x \frac{N_B(t)}{t^2}\,dt \\
&\geqslant \gamma\cdot \Biggl(\frac{N_S(x)}{x} + \int_1^x \frac{N_S(t)}{t^2}\,dt\Biggr) - \int_1^{x_{\gamma}} \frac{\lvert \gamma N_S(t) - N_B(t)\rvert}{t^2}\,dt \\
&= \gamma\cdot L_S(x) - \int_1^{x_{\gamma}} \frac{\lvert \gamma N_S(t) - N_B(t)\rvert}{t^2}\,dt \\
&\geqslant \gamma\cdot L_S(x) - \log x_{\gamma}
\end{align}
and consequently
$$\liminf_{x \to \infty} \frac{L_B(x)}{L_S(x)} \geqslant \gamma\,.$$
(This uses $\lvert\gamma N_S(t) - N_B(t)\rvert \leqslant t$, which follows from $0 \leqslant N_B(t) \leqslant t$ and $0 \leqslant \gamma N_S(t) \leqslant \gamma t \leqslant t$ [clearly $c \leqslant 1$, hence $\gamma < 1$] and $\frac{\log x_{\gamma}}{L_S(x)} \to 0$.) Since $\gamma < c$ was arbitrary the left inequality of $(\ast)$ follows. The middle inequality follows from the definitions of $\liminf$ and $\limsup$, and the right inequality of $(\ast)$ is proved similar to the left. Let $\Gamma > C$. Then there is an $x_{\Gamma}$ such that $N_B(x) \leqslant \Gamma\cdot N_S(x)$ for $x \geqslant x_{\Gamma}$. Essentially the same computation, just with the sense of the inequalities reversed, $\gamma$ replaced with $\Gamma$, the last integral added instead of subtracted, yields
$$\limsup_{x \to \infty} \frac{L_B(x)}{L_S(x)} \leqslant \Gamma\,.$$
Again, since this holds for all $\Gamma > C$, the right inequality of $(\ast)$ follows.
Concerning your second "aside",
Also in the link provided above, in order to show that the sum of reciprocals of a set of primes having positive relative density $\alpha > 0$ is divergent, the accepted answer goes through logarithmic density, however I feel that there should be more elementary arguments for the same.
this is a bit delicate. I sketched an example of a set $A$ of primes having upper (relative) natural density $1$ in the set of all primes such that the series of the reciprocals of the primes in $A$ converges. Thus you should not expect a very simple argument. On the other hand, as I also mentioned over there, we don't need a (relative) natural density, a positive lower (relative) natural density suffices. The proof is above: Abel summation yields
$$L_B(x) \geqslant \gamma L_S(x) - \log x_{\gamma}\,,$$
and the right hand side tends to $\infty$ for $x \to \infty$. The argument is fairly elementary, I'd say, but of course not totally trivial. I can't think of a more elementary argument.
Best Answer
What do you get with $$\bigcup_{k\ge 1} \{ p \text{ prime}, p\in [e^{e^k},2 e^{e^k}]\}$$