Denoting by $A$ the complement of $P$ in $[1,100]$, you are asking how large can $A$ be given that it does not contain any $10$-term arithmetic progression. The more general question
How large can a set $A\subset[1,N]$ be given that it does not contain any $k$-term arithmetic progression?
is one of the central problems in combinatorial number theory (additive combinatorics). There is no chance to give a precise answer, as an "explicit" function of $N$ and $k$; however, in your special situation, where $N=n^2$ and $k=n$, this might be possible.
Here is, at least, a proof-of-concept computation showing that if $P\subset[1,n^2]$ blocks every $n$-term progression, then $|P|>1.5n-0.5$ (which is half-way between the trivial estimate $|P|>n$, and the estimate $|P|>2(n-1)$ you are asking about). I strongly suspect that with some further effort, it can be improved to something like $|P|>2n-O(1)$.
Write $P=\{p_1,\dotsc,p_K\}$ where $K=|P|$ and $1\le p_1<\dotsb<p_K\le n^2$. Without loss of generality, we assume $p_1=1$ and $p_K\ge n^2-n$. For any $d\in[1,n]$, the set $P$ should contain an element form each residue class modulo $d$. It follows that there are at most $K-d$ indices $i\in[1,K-1]$ such that $p_{i+1}-p_i=d$. Also, if $d>n$, then there are no such indices at all. Denoting the number of the indices in question by $\nu(d)$, we thus have $$ (n^2-n) - 1 \le p_K-p_1 = \sum_{d=1}^n \nu(d) \le \sum_{d=1}^n (K-d) = Kn - \frac12\,n(n+1), $$ implying $|P|=K>1.5n-0.5$.