No, this is not always possible. Even if $X$ happens to be normal.
Note that the set $\mathbb{C} \setminus \{ 0 \} \subseteq \mathbb{C}$ may be represented as $\bigcup_{n \in \mathbb{N}} \{ z \in \mathbb{C} : |z| \geq \frac{1}{n} \}$. So in addition to being open, $\mathbb{C} \setminus \{ 0 \}$ is an Fσ-set. One can show that if $f : X \to Y$ is continuous (arbitrary spaces $X$ and $Y$), then $f^{-1} [ B ]$ is Fσ in $X$ whenever $B$ is Fσ in $Y$. So for any space $X$ and any continuous mapping $f : X \to \mathbb{C}$ the set $f^{-1} [ \mathbb{C} \setminus \{ 0 \} ]$ is an open Fσ subset of $X$.
A space $X$ in which every open subset is Fσ is sometimes called a perfect space, and a normal perfect space is called perfectly normal. Perfectly normal spaces have the property that for any open subset $U \subseteq X$ there is a continuous $f : X \to \mathbb{R}$ such that $f^{-1}[ \mathbb{R} \setminus \{ 0 \} ] = U$. (The same obviously holds for complex-valued continuous functions.)
And there are certainly examples of Hausdorff (even normal) spaces which are not perfect. One is the ordinal space $[0 , \omega_1 ] = \omega_1+1$. Here the interval $[0,\omega_1)$ is open, but is not Fσ. (Any closed subset which does not contain $\omega_1$ is countable, and the union of countably many countable sets is countable.) So there is no continuous function $f : [0,\omega_1] \to \mathbb{C}$ such that $[0,\omega_1) = f^{-1} [ \mathbb{C} \setminus \{ 0 \} ]$.