Because $f$ is linear, the minima and maxima over a particular interval of $f$ occur at its endpoints; in this case, $0$ and $1$. From $\vert f(1)\vert, \vert f(0)\vert\leq 1$, we have $1\geq b\geq -1$ and $1\geq a+b\geq -1$. Then \[S=(a+1)(b+1)=ab+a+b+1\leq 0.25(a+b)^2+(a+b)+1 \leq 2.25.\]Because $b+1$ is always nonnegative, it suffices to make $a+1$ as small as possible to find the minimum. From $a+1\geq a+b\geq -1$ we get $(a+1)(b+1)\geq -1\cdot 2 = -2$. The range of values of $S$ is $\boxed{[-2, 2.25]}$. $S=2.25$ is satisfied when $a=b=\frac{1}{2}$. $S=-2$ is satisfied when $a=-2, b=1$.