False. Let the quadratic be $(x-0.2)(x-0.3)=x^2-0.5x+0.06$. Clearly at least one root lies in the interval $[-1,1]$, and both roots are distinct and real, however none of the roots lie in $(-b,b)$.

Corrected. Exactly one root lies in [-1,1].

Solution

Let $r_1, r_2$ be the roots of $x^2+ax+b$ with $r_1$ in $[-1;1]$. Clearly if $b>1$ then $r_1$ also lies in $(-b; b)$. Suppose $b = 1$. Then, by Vieta's formulas, $r_1r_2=1$. If $-1<r_1<1$, then $r_1$ lies in $(-b; b)$. If $r_1=\pm1$, then $r_2=r_1$ (so their product is $1$), contradicting the roots are distinct. We finalize by checking $b<1$. Suppose $r_1$ is not in $(-b;b)$. We tackle two cases: $r_1$ positive and $r_1$ negative. Note that $r_1$ and $r_2$ have the same sign, as they multiply to a positve number. Hence, in the first case we have $b<r_1<1$, so $br_2<r_1r_2=b$, implying that $0<r_2<1$. However, this contradicts that only one of $r_1,r_2$ lies in $[-1;1]$. Similarly, in the second case we have $b<-r_1<1$, so $br_2>-r_1r_2=-b$, implying that $-1<r_2<0$, which again contradicts that only one of $r_1, r_2$ lies in $[-1;1]$. Therefore, if $b<1$, $r_1$ must lie in $(-b;b)$. We have shown in every case that $r_1$ lies in the interval $(-b;b)$, as desired. $\blacksquare$