It is hard to find such $a$. Indeed, according to Wikipedia, the set of such $a$ is countable. (Remark that Corollary 3 of Zeldich’s attraction theorem states that a set $\{a\in (1,\infty): \{a^n\}\mbox{ is not dense in }[0,1]\}$ is meager in $(1,\infty)$.)
Moreover, if the answer to a longstanding Pisot-Vijayaraghavan problem is affirmative then there is no such $a$. Indeed, for a contradiction pick such $a$. The affirmative answer to Pisot-Vijayaraghavan problem implies that $a$ is a Pisot–Vijayaraghavan number. Let $a_1=a$ and $a_2, \dots, a_m$ be the conjugates of $a$. Then for each natural $n$, $\sum_{k=1}^m a_k^n$ is integer.
For each $k=1,\dots, m$ let $a_k=r_ke^{\varphi_ki}$ for some positive $r_k$ and $\varphi_k$. Let $G=\Bbb T^m=\{z\in\Bbb C:|z|=1\}$ be a (multiplicative) topological group and $g=(e^{\varphi_1i}, e^{\varphi_2i},\dots, e^{\varphi_m i})$ be an element of $G$. Since the group $G$ is compact, it is well-known and easy to show that $G$ topologically periodic, so for any neighborhood $U$ of the identity of $G$ there exists arbitrarily big natural number $n$ such that $g^n\in U$. Pick $U_0=\{(x_1,\dots,x_m)\in\Bbb T^m: \forall i (\operatorname{Re} x_i\ge 0)\}$.
Let $N>0$ be any number. Since $a$ is a Pisot-Vijayaraghavan number, we have $r_k<1$ for each $k=2,\dots, m$. So there exists $M>0$ such that $\sum_{k=2}^m r_k^M<1/2$. Pick $n>N,M$ such that $g^n\in U_0$. Then for some integer $K$ we have
$$K=\sum_{k=1}^m a_k^n=\sum_{k=1}^m r_k^ne^{n\varphi_ki} =\sum_{k=1}^m \operatorname{Re} r_k^n (e^{n\varphi_ki})=
a^n+\sum_{k=2}^m r_k^n \operatorname{Re} e^{n\varphi_ki}.$$
So $a^n\le K\le a^n+\sum_{k=2}^m r_k^n\le a^n+\sum_{k=2}^m r_k^M<a^n+1/2$. Thus $\{a^n\}>1/2$, a contradiction.