The following proof shows that the almost independent set problem is hard using a Cook reduction.
It is well-known that approximating the independent set problem is hard within every polynomial factor [1].
Suppose that if there is a polynomial algorithm for the almost independent set problem, then there is an $x-100$-approximation (i.e., the algorithm returns an independent set of size larger than or equal to $n_{opt}-100$, where $n_{opt}$ is the size of the optimal independent set) polynomial algorithm for the independent set problem. Thus, the problem is NP-hard.
The approximation algorithm for the independent set problem goes as follows:
1) Derive the largest almost independent set $S$.
2) Remove from $S$ all vertices that share the same edge. Denote the result by $S'$
3) Return $S'$.
The algorithm runs in polynomial time.
Correctness: Let $S_{opt}$ be the largest independent set. Then
$$|S'|\geq |S|-100\geq |S_{opt}|-100.$$
The first inequity follows as remove up to 100 vertices.
The second inequity follows as every independent set is also an almost independent set, and thus the size of the largest independent set is smaller than or equal to the size of the largest almost-independent set.
Thus, the algorithm is a $x-100$ approximation for the independent set problem, and unless P=NP, there is no polynomial algorithm for the almost independent set problem.
[1]-https://en.wikipedia.org/wiki/Independent_set_(graph_theory)