The correct statement is given by the deterministic Miller-Rabin test coupled with an estimate of Bach under GRH.
Theorem. Assume GRH. Let $n$ be an odd integer, and write $n-1$ as $2^st$ with $t$ odd. Then $n$ is composite if and only if there is $1<b<2\log^2 n$ such that $b^t\not\equiv 1\pmod{n}$ and $b^{2^r t}\not\equiv -1\pmod{n}$ for all $0\leq r<s$.
For more details see Section V.1 in Koblitz: A course in number theory and cryptography (2nd edition, GTM 114, Springer, 1994).
Added 1. The OP wanted more specific answers to his questions Q1 and Q2. Here they are:
Q1: The quoted paragraph without the parentheses gives a complete proof for the square-free test, so Granville's paper can be taken as a reference (although I am sure Lenstra wrote this down earlier in some form). The part in parentheses is false, so there is no reference for it. A corrected version for this part is given above, and a reference for it is Koblitz's book (or the original papers by Miller, Rabin, Bach).
Q2: There is nothing wrong with the quoted counterexample. It is a genuine counterexample to the part in parentheses, which shows in particular that the part in parentheses is false.
Added 2. A more complete exposition of this material can be found in Sections 10.2 and 10.5 of Shoup: A computational introduction to number theory and algebra (version 2).
Added 3. Using Theorem 1.1 of Lamzouri-Li-Soundararajan in place of Theorem 2 of Bach, we get the following stronger version of the above theorem.
Theorem. Assume GRH. Let $n>3000$ be an odd integer, and write $n-1$ as $2^st$ with $t$ odd. Then $n$ is composite if and only if there is $1<b<\log^2 n$ such that either $b\mid n$, or $b^t\not\equiv 1\pmod{n}$ and $b^{2^r t}\not\equiv -1\pmod{n}$ for all $0\leq r<s$.
Added 4. In deducing the above theorems it is good to have in mind the following supplement to Theorem 10.3 in Shoup's book: if $n$ is composite, then $L_n'$ generates a proper subgroup of $\mathbb{Z}_n^\times$. Indeed, this supplement follows by the proof of the quoted theorem.