Let $\mathbb{D}^*=\{z \in \mathbb{C} \ | \ 0 < |z| < 1 \}$ denote the unit punctured disk in the complex plane. Riemann's theorem about removable singularities in particular implies the following:
If $f:\mathbb{D}^* \to \mathbb{C}$ is a bounded holomorphic function, then there exists a unique holomorphic extension $F: \mathbb{D} \to \mathbb{C}$ to the whole unit disk.
Question is it possible to get an analogous extension statement if we replace the punctured unit disk by a complex annulus $A(r,R)=\{ z \in \mathbb{C} \ | \ r < |z| < R \}$? In this case "remove the singularity" should mean something like being at least continuous up to the [inner] boundary of the annulus, or maybe even holomorphic on some open neighborhood of it. More explicitly if $f:A(r,R)\to \mathbb{C}$ is bounded and holomorphic, is there some hope to get some continuous extension to $\{ z \in \mathbb{C} \ | \ r \leq |z| < R \}$, or maybe some holomorphic extension to $\{ z \in \mathbb{C} \ | \ r - \varepsilon < |z| < R \}$?
No, there is no such hope. For a simple example, take $f(z)=1/z$.
In fact, an annulus is a domain of existence for $H^\infty$ as well as $A$, i.e. there exists bounded resp. continuous up to the boundary holomorphic functions on the annulus which can't be extended holomorphically across any boundary point.
More concretely, let $g$ be your favorite bounded/continous up to the boundary holomorphic function on the unit disc that can't be extended across any boundary point, and let $f(z) = g(z)+g(r/z)$. Then $f$ can't be extended across any boundary point of $r<|z|<1$.