Solving Exact2IS using IS
First of all, if you can determine whether a graph $G$ contains an independent set of size $k$, then you can also find such a set efficiently. This is known as "search-to-decision reduction". Here is the basic idea. Choose an arbitrary vertex $v$, and remove it. If the graph still has an independent set of size $k$, then keep going. Otherwise, all independent sets of size $k$ contain $v$. Accordingly, remove $v$ and all its neighbors, and find an independent set of size $k-1$ in the remaining graph. In this way, you can recover an independent set of size $k$.

Second, how to check whether a graph contains at least two independent sets of size $k$, assuming $k \geq 2$. First, you determine whether in contains at least one. Suppose that it does, say $I$. If $J$ is any other independent set, then $I \setminus J$ and $J \setminus I$ are both nonempty (since $|I| = |J|$). In particular, if $x \in I \setminus J$ and $y$ is any other vertex in $I$, then even after adding the edge $(x,y)$, the set $J$ will constitute an independent set.

This leads to the following algorithm: for each pair of vertices $x,y \in I$, check whether $G + (x,y)$ contains an independent set of size $k$. If so, this independent set is necessarily different from $I$. Conversely, if an independent set of size $k$ different than $I$ exists, then necessarily it will be an independent set in $G + (x,y)$ for some $x,y \in I$.

Third, how to check whether a graph contains exactly two independent sets of size $k$. This is the same as checking whether a graph contains at least three independent sets. I think that at this point, it's better if you try to generalize the above argument from two to three independent sets. You might get stuck, but you won't know until you try.