Let $D\subset\mathbb C$ be a domain, $S\subset D$ finite, $p\in S$ and $r,R>0$ such that $K_{r,R}(p)\subset D$, where $K_{r,R}(p)=\{z\in\mathbb C\mid r<|z-p|<R\}$. Also let $f:D\backslash S\to\mathbb C$ be holomorphic.
Is $p$ an essential singularity if for the Laurent series $f(z)=\sum_{n\in\mathbb Z}a_n z^n$ on $K_{r,R}(p)$ we have $a_n\neq 0$ for all $n<0$?
I know that this statement is true if $f$ has a Laurent series on an annulus $0<|z-p|< R$ which is not the case. Let's take $p=1$ and $f(z)=\frac{-1}{1-z}$, so $p$ is a pole of order 1 and on $|z|>1$ we have $f(z)=\sum_{n=-\infty}^{-1}z^n$. Does this work as a counter example or am I missing something? I take $D=\overline{K_{0,2}(1)}$ and $S=\{1\}$.
Your counter-example is not a counter-example, for the reason you provided yourself in the comments.
Define $f\colon\mathbb{C}\setminus\{0,1\}\longrightarrow\mathbb C$ defined by $f(z)=\frac1{z-z^2}$. Clearly, $f$ has no essential singularity at $0$. However, its Laurent series in the region $|z|>1$ is$$-z^{-2}-z^{-3}-z^{-4}-\cdots$$
You might say that this is not a counter example, since $a_{-1}=0$. But then consider $g(z)=\frac{z+1}{z-z^2}$. Its Laurent Series in the same region is$$-z^{-1}-2z^{-2}-2z^{-3}-2z^{-4}-\cdots$$
However, there is a mistake within the question itself: when you wrote $f(z)=\sum_{n\in\mathbb Z}a_nz^n$, what you meant was $f(z)=\sum_{n\in\mathbb Z}a_n(z-p)^n$.