Can we have a power density but not a natural density?

Yes, we can!

For $k \in \mathbb{N}$, let $a_k = \bigl((2k-1)!\bigr)^2$, $b_k = \bigl((2k)!\bigr)^2$ and $c_k = (1 + \varepsilon_k)a_k$, where $\varepsilon_1 = 1$ and $\varepsilon_k = \frac{1}{k\log k}$ for $k > 1$, and

$$B = \bigcup_{k = 1}^{\infty} [a_k, b_k),\quad A = \bigcup_{k = 1}^{\infty} [a_k, c_k)\,.$$

Then $A$ - and a fortiori $B$ - is substantial:

$$\sum_{k = 1}^K \Biggl(\sum_{a_k \leqslant n < c_k} \frac{1}{n}\Biggr) > \sum_{k = 1}^K \log (1 + \varepsilon_k) > \frac{1}{2}\sum_{k = 1}^K \varepsilon_k > \frac{1}{4} \log \log K\,.$$

$A$ has no natural density in $B$:

For $k \geqslant 2$ we have $S_{A,0}(c_k) \geqslant \varepsilon_k a_k$ and $S_{B,0}(c_k) \leqslant b_{k-1} + \varepsilon_k a_k < \frac{k+\log k}{k^2\log k} a_k$, so

$$\frac{S_{A,0}(c_k)}{S_{B,0}(c_k)} \geqslant \frac{k}{k+\log k}$$

and hence $\overline{D}_0(A;B) = 1$. But $\underline{D}_0(A;B) = 0$ since

$$\frac{S_{A,0}(b_k)}{S_{B,0}(b_k)} \leqslant \frac{(1+\varepsilon_k)a_k}{b_k - a_k} = \frac{1+\varepsilon_k}{4k^2-1}\,.$$

And we have $D_{\frac{1}{2}}(A;B) = 0$ since for $a_{k+1} \leqslant m < a_{k+2}$

$$S_{B,\frac{1}{2}}(m) \geqslant \sum_{n = a_{k}}^{b_{k}-1} \frac{1}{\sqrt{n}} > 2\sum_{n = a_k}^{b_k-1} \frac{1}{\sqrt{n+1} + \sqrt{n}} = 2(\sqrt{b_k} - \sqrt{a_k}) = (4k-2)\cdot (2k-1)! \geqslant (2k)!$$

and

$$S_{A,\frac{1}{2}}(m) \leqslant \sum_{n = 1}^{c_k-1} \frac{1}{\sqrt{n}} + \sum_{n = a_{k+1}}^{c_{k+1}-1} \frac{1}{\sqrt{n}} < 2\sqrt{c_k} + \varepsilon_{k+1} \sqrt{a_{k+1}} < 3(2k-1)! + \frac{(2k+1)!}{(k+1)\log (k+1)}$$

whence

$$\frac{S_{A,\frac{1}{2}}(m)}{S_{B,\frac{1}{2}}(m)} < \frac{3}{2k} + \frac{2}{\log (k+1)}\,.$$

It is not hard to modify the example to get $0 < D_{\frac{1}{2}}(A';B) < 1$.