Denote $a=-b\in(0,2)$.
$P(x)=(x+1)^n(x^2-ax+1)=\sum_{k=0}^{n+2}T_kx^k$.
Using $(x+1)^n=\sum_{m=0}^nC_n^mx^m$, results:
$T_0=T_{n+2}=1$;
$T_1=T_{n+1}=n-a>0, \forall n\ge2$;
$T_k=C_n^k-aC_n^{k-1}+C_n^{k-2}$ for $2\le k\le n$.
$C_n^k-aC_n^{k-1}+C_n^{k-2}=\dfrac{n!}{k!(n-k)!}-a\cdot\dfrac{n!}{(k-1)!(n-k+1)!}+\dfrac{n!}{(k-2)!(n-k+2)!}=$
$=\dfrac{n!}{k!(n-k+2)!}\cdot[(a+2)k^2-(a+2)(n+2)k+(n+1)(n+2)]$.
The second degree polynomial with the variable $k$
$Q(k)=(a+2)k^2-(a+2)(n+2)k+(n+1)(n+2)$ attains the minimum value for
$k_0=\dfrac{n+2}{2}$.
$Q(k)\ge Q(k_0)=\dfrac{(n+2)[(2-a)n-2a]}{4}$.
$a<2\Longrightarrow 2-a>0$.
$Q(k_0)>0\Longleftrightarrow n>\dfrac{2a}{2-a}$.
Hence, for $n>\dfrac{2a}{2-a}$:
$T_k=\dfrac{n!}{k!(n-k+2)!}\cdot Q(k)>0, \forall k\in\{2,3,\dots,n\}$.
Results for $n>\max\left\{1;\dfrac{2a}{2-a}\right\}$:
$T_k>0, \forall k\in\{0,1,2,\dots,n+2\}$.