We claim the answer is $8$. Let us first find the points in a connected component. Let $l_P$ be the line passing through a point $P$ and parallel to $AB$. Let $l'_P$ be the reflection of $l_P$ with respect to $AB$. First note that the reflection of any point $P$ with respect to each of $A$ and $B$ lies on $l'_P$. Hence, any point connected to $P$ lies on $l_P$ or $l'_P$.

We claim if $P$ is connected to a point $Q$, then $Q$ can be attained by some number of transformations by $\pm 2\overrightarrow{AB}$ and at most one reflection with respect to $A$: the combination of any two reflections with respect to $A$, $B$ is a transformation.

Transforming by $\pm 2\overrightarrow{AB}$ does not change the parity of coordinates, so if we define an equivalence relation $P \equiv Q$ if and only if $P = Q + 2n\overrightarrow{AB}$, $n \in \mathbb{Z}$, points will form at least $14$ classes. Every connected component has points from at most two classes. Note that $A$ and $B$ are not connected and their components consist of exactly one class, so we have at least $8$ components.

As an example, which is easy to verify, take the two points having only half a unit distance from the center of the table as $A$, $B$.