Zeros and poles of rational function.

Just because $s = 0$ is a zero of $a(s)$ does not mean $s = 0$ is a zero of $f(s)$. To be a zero, the numerator must be zero and the denominator not (otherwise, apply l'Hopital's rule to agonizingly cancel the common factor in $a$ and $b$). Happily, $b(0) neq 0$ and once you observe this fact, you have found all the zeroes of $f$ (and shown that it is a genuine zero of $f$). (Example: $z/z$ has a removable singularity at $z = 0$ and upon removal takes the value $1$.)

$f$ has a pole at $s$ if $b(s) = 0$ and $a(s) neq 0$. With the above modification, you would have already disposed of the possibility of common zeroes in $a$ and $b$, so every zero of $b$ is a pole, as you write.