This is a revised version of my original answer. I fixed some inconsistencies and made the text more readable.
The correct statement is given by the deterministic Miller-Rabin test coupled with an estimate of Bach under GRH. Let us follow Sections 10.2 and 10.5 of Shoup: A computational introduction to number theory and algebra (version 2). Let $n$ be an odd integer, and write $n-1$ as $2^st$ with $t$ odd. Consider
$$ L_n':=\left\{a\in\mathbb{Z}_n^\times:\ a^{2^s t}=1\quad\text{and}\quad a^{2^{r+1}t}=1\Longrightarrow a^{2^rt}=\pm 1\quad\text{for}\quad 0\leq r<s \right\} $$
For $n$ prime, $L_n'$ equals $\mathbb{Z}_n^\times$. For $n$ composite, the proof of Theorem 10.3 in Shoup's book shows that $L_n'$ generates a proper subgroup of $\mathbb{Z}_n^\times$. Combining these statements with Theorem 2 of Bach, we get:
Theorem 1. Assume GRH. Let $n$ be an odd integer. Then $n$ is composite if and only if there is $1<b<2\log^2 n$ such that $b$ mod $n$ lies outside $L_n'$.
Using Theorem 1.1 of Lamzouri-Li-Soundararajan in place of Bach's result, we get the following stronger version:
Theorem 2. Assume GRH. Let $n>3000$ be an odd integer. Then $n$ is composite if and only if there is $1<b<\log^2 n$ such that either $b\mid n$, or $b$ mod $n$ lies outside $L_n'$.
Added. 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 Shoup's book.
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.