I claim all such pairs are $(a,0)$ where $0<a<4$. Take $x_0$ such that $f(x_0)=0$. Since $f(f(x_0))=f(0) = 0$ as well, we have $b=0$ and $f(x)=x^2+ax=x(x+a)$ and $f(f(x)) = x(x+a)(x^2+ax+a)$. Thus, we want to ensure $x^2+ax+a$ has no real solutions. This is true iff its discriminant is negative, i.e., $a^2<4a$. So, $(a-2)^2 <4$, forcing $-2<a-2<2$, namely $a\in(0,4)$.