Basic definitions:
Upper and lower asymptotic density (a.k.a. natural density):
$$\overline d(A)=\limsup \frac{A(n)}n$$
$$\underline d(A)=\liminf \frac{A(n)}n$$
Upper and lower uniform density (a.k.a. Banach density):
$$\overline u(A)=\lim_{s\to\infty} \max_{t\ge 0}\frac{A(t+1,t+s)}{s}$$
$$\underline u(A)=\lim_{s\to\infty} \min_{t\ge 0}\frac{A(t+1,t+s)}{s}$$
where $A(m,k)=|A\cap\{m,m+1,\dots,k\}|$ and $A(n)=A(1,n)$.
Note that $\underline d(\mathbb N\setminus A)=1-\overline d(A)$ and $\underline u(\mathbb N\setminus A)=1-\overline u(A)$.
It is known that $\underline u(A)\le \underline d(A) \le \overline d(A) \le \overline u(A)$, see e.g. [GLS].
The condition that $d_n$ is unbounded is equivalent to $\underline u(A)=0$.
(If $d_n$ is unbounded that we can found arbitrarily large $s$ with $A(t+1,t+s)=0$. On the other hand, if $d_n\le M$, then $A(t+1,t+s) \ge \left\lfloor\frac{s}M\right\rfloor$ and $\underline u(A)\ge \frac 1M$.)
It is known that $\overline u(A)=\sup\{\mu(A); \mu\text{ is a shift-invariant mean on }\mathbb N\}$. The proof of this fact can be found in [B]. (EDIT: Now I realized that you have seen this result in an answer to another your question Invariant means on the integers Both [B] and this answer concern invariant means on $\mathbb Z$ and not $\mathbb N$, but this should not make much difference.)
Knowing all of this, you in fact ask whether $\underline u(A)=0$ implies $\overline d(A)=0$ or even $\overline u(A)=0$. Many counterexamples can be found. Just one of them: For the set
$$A=\mathbb N\setminus \bigcup_{k=1}^\infty \{10^k+1,\dots,10^k+k\}$$
we get $\underline u(A)=0$ and $\underline d(A)=\overline d(A)=\overline u(A)=1$.
(If I remember correctly, I have seen the result that for any choice of $0\le a \le b \le c \le d\le 1$ there exists a set $A$ such that the values of $\underline u(A)$, $\underline d(A)$, $\overline d(A)$, $\overline u(A)$ are $a$, $b$, $c$ and $d$, respectively; but I might be mistaken and I cannot find any reference right now. Maybe I mixed it up with a similar result for some other type of densities.)
EDIT: The question whether the result mentioned in preceding paragraph holds is now posted on MO as a separate question: On the independence of lower and upper asymptotic and Banach densities
Perhaps I should mention that several equivalent definitions of Banach/uniform density appear in the literature. They are compared e.g. in [GTT]. (E.g. [B] works with a different - but equivalent - definition.)
Some references to papers where this notion was studied under the name uniform density are given in [GLS]. If I am not mistaken, the term Banach density was coined by Furstenberg [F].
[B] Mathias Beiglbock: An ultrafilter approach to Jin’s Theorem
http://www.mat.univie.ac.at/~mathias/UltraJin_final.pdf
[F] H. Furstenberg. Recurrence in ergodic theory and combinatorial number theory. Princeton University Press, Princeton, 1981.
[GLS] Z. Gáliková, B. Lászlo and T. Šalát: Remarks on uniform density of sets of integers
http://www.emis.ams.org/journals/AMI/2002/acta2002-galikova-laszlo-salat.pdf
[GTT] Georges Grekos, Vladimír Toma and Jana Tomanová. A note on uniform or Banach density.
http://ambp.cedram.org/ambp-bin/fitem?id=AMBP_2010__17_1_153_0
Best Answer
I think that it is fairly straightforward to get such a set. You can simply get the set as a union of intervals:$\newcommand{\intrvl}[2]{\langle{#1},#2)}\newcommand{\intrvr}[2]{({#1},#2\rangle}\newcommand{\limti}[1]{\lim\limits_{{#1}\to\infty}}$ $$A=\bigcup\limits_{n\in\omega} \intrvr{a_n}{b_n}.$$ And you choose the intervals in a suitable way. You want them long enough to get $d_{uB}(A)=1$. And, at the same time, you want start each interval far enough to keep $d_u(A)=0$. So it suffices to choose them in such way that \begin{align*} \limti n (b_n-a_n)=\infty\\ \limti n \frac{\sum_{k=1}^n (b_k-a_k)}{b_n}=0 \end{align*} The first conditions will help you to get upper Banach density (a.k.a. upper uniform density) equal to one. And the second condition will keep the asymptotic density (natural density) zero.
For example, you can take $a_n=2^n$ and $b_n=2^n+n$.
Several posts on this site are a bit related - and the answers include some references which might be useful: