Let $d(A)$ denote the natural density of $A\subset \mathbb{N}.$
If $0 < \alpha < 1\ $ and $\ d(A) = \alpha,\ $ then is this enough to imply
$$ \lim_{ N\to\infty } \left( \frac{ \displaystyle\sum_{n\in A}^{n\leq N} \frac{1}{n} }{ \displaystyle \sum_{n\in \mathbb{N}}^{n\leq N} \frac{1}{n} } \right) \equiv \lim_{ N\to\infty } \left( \frac{ \displaystyle\sum_{n\in A}^{n\leq N} \frac{1}{n} }{ \log(N) } \right) = \alpha\ ? $$
I think it should be "obviously" true if, for example, $\ A = \lbrace{ \left\lceil \frac{n}{\alpha} \right\rceil: n\in\mathbb{N} \rbrace},\ $ but to me it is not so obvious for general $A.$
I was hoping to use Stolz–Cesàro, but don't quite see how. Or maybe by looking at lots of intervals like $[N, 1.01N]$ for very large $N,$ but I don't have anything concrete yet. Or maybe there is a counter-example.
Edit: maybe the result is (also?) true for $\ \alpha=0\ $ and/or $\ \alpha = 1.$
Yes, your statement is true. Abel's partial summation formula states that
$$\sum_{n\le N}{a_n f(n)} = A(N)f(N)\ - \int_1^N{A(t)f'(t)\ dt}$$
where $A(n) := \sum_{k\le n}{a_k}$. This is useful if you already know the partial sums of $a_n$, but need to find the partial sums of $a_n$ weighted by some function $f$. If we let $a_n$ be the indicator function defined by
$$a_n := \begin{cases} 1 &\text{if } n\in A\\ 0 &\text{Otherwise}\end{cases}$$
and another function $S(n) = \sum_{k\le n}{\frac{a_k}{k}}$, we see that your limit is $S(N)/\log(N)$ and our goal is to show that this approaches $\alpha$. By your assumption $d(A) = \alpha$, we know that $\lim_{N\to\infty}{\frac{A(N)}{N}} = \alpha$. To find $S(N)$, we simply apply the partial summation formula as follows with $f(n)=1/n$ -
$$S(N) = \frac{A(N)}{N}\ - \int_1^N{A(t)\left(-\frac{1}{t^2}\right)\ dt}$$
$$= \frac{A(N)}{N}\ + \int_1^N{\frac{A(t)}{t^2}\ dt}$$
By definition, we know that $A(N)/N$ is bounded by $1$, and will approach zero when divided by $\log(N)$. Furthermore, the integral term can be expressed as so
$$\int_1^N{\frac{A(t)}{t^2}\ dt} = \int_1^N{\frac{\alpha t}{t^2}\ dt} + \int_1^N{\frac{A(t) -\alpha t}{t^2}\ dt}$$
$$= \alpha\log(N) + E(N)$$
This means that for $S(N)/\log(N)$ to approach $\alpha$, it remains to show that $E(N)/\log(N)$ tends to zero. Fix $\epsilon > 0$, and let $N$ be large. By the limit $A(n)/n \to \alpha$, there exists an $n_0(\epsilon)$ such that $\left|\frac{A(n)}{n}-\alpha\right|\lt\epsilon$ for all $n\ge n_0$. Or equivalently, $\left|A(n)-\alpha n\right|\lt\epsilon\cdot n$. Thus, we have
$$|E(N)| \le \int_1^N{\frac{|A(t) -\alpha t|}{t^2}\ dt} = \int_1^{n_0}{\frac{|A(t) -\alpha t|}{t^2}\ dt} + \int_{n_0}^N{\frac{|A(t) -\alpha t|}{t^2}\ dt}$$
$$< \int_1^{n_0}{\max\left(\frac{|A(t) -\alpha t|}{t^2}\right)\ dt} + \int_{n_0}^N{\frac{\epsilon\cdot t}{t^2}\ dt}$$
$$= M\cdot n_0(\epsilon) + \int_{n_0}^N{\frac{\epsilon}{t}\ dt} = c_{\epsilon} + \epsilon\log(N)$$
where $c_{\epsilon}$ is a constant only depending on $\epsilon$. Finally, since $\epsilon$ is arbitrary, we have that
$$\limsup_{N\to\infty}{\frac{|E(N)|}{\log(N)}} = 0,$$
which completes the proof.
Remarks:
The natural density $d(A)$ is an example of the average order or "mean value" of a function. In this case, the function is just an indicator of whether or not a number $n$ is in $A$, but in general, this can be any arithmetic function $f$. More formally, the mean value is defined as
$$M(f) = \lim_{N\to\infty}{\frac{1}{N}\sum_{n\le N}{f(n)}}$$
The logarithmic mean value is also defined as
$$L(f) = \lim_{N\to\infty}{\frac{1}{\log(N)}\sum_{n\le N}{\frac{f(n)}{n}}}$$
In general, one can show that the equality $M(f)=\alpha\ $ implies that $L(f)$ exists and is also equal to $\alpha$. This can be proved using the same steps as shown above. In fact, $\alpha$ can take on any complex value, of course this only applies to mean values and not densities of sets.