整理$\sqrt {x-2010}+y^{2}=-2y-1$得：

$\sqrt {x-2010}+\left(y+1\right)^{2}=0$

$\therefore \sqrt {x-2010}=0$，$\left(y+1\right)^{2}=0$

$\therefore x=2010$，$y=-1$，

$\therefore x+y+xy=2010-1-2010=-1$.

故选$A$.