The number $z = i/\sqrt2$ seems to work!

Given $x \in [-2/3,4/3]$ we can find a "negabinary" expansion
$$x = \sum_{k=0}^\infty (-1)^k\frac{b_k}{2^k},$$ where each $b_k \in \{0,1\}$.
Similarly, given $y \in [-2/3\sqrt2,4/3\sqrt2]$ we can find
$$y = \frac{1}{\sqrt2}\sum_{j=0}^\infty (-1)^j\frac{c_j}{2^j},$$
where each $c_j \in \{0,1\}$.
Therefore,
$x + iy = \sum_{n \in A} z^n$ where
$$A = \{2k : b_k = 1\} \cup \{2j+1 : c_j = 1 \}.$$

As explained in comment contributions by Bart Michels, Jochen Wegenroth and myself, $|z| \geq 1/\sqrt2$ is necessary. By definition, $$\Sigma_z = z\Sigma_z\cup(1+z\Sigma_z).$$
If $\mu$ denotes Lebesgue measure, then $\mu(z\Sigma_z) = |z|^2\mu(\Sigma_z)$ thus $\mu(\Sigma_z) \leq 2|z|^2\mu(\Sigma_z)$. Since $\Sigma_z$ is compact, it has finite measure and thus if $\mu(\Sigma_z)>0$ then we must have $|z|^2 \geq 1/2$.

It remains open whether $|z|\geq1/\sqrt2$ and $z \notin \mathbb{R}$ is sufficient for $\Sigma_z$ to contain $0$ in its interior.