The observation regarding the logarithm shows that the product exists if $s >2$, since $\ln(1+x) < x$ for $x >0$, so that the expression for $\ln(A)$ is less than $\sum_{p \in P} (p-1)^{1-s}$, which converges. However, for $1 < s < \leq 2$, the product diverges, since, for a given $p$, the contribution to the product from $p$ is at least $1 + (p-1)^{1-s}$, (using the binomial theorem), so at least $\frac{p}{p-1}$. Hence the product is at least $\frac{1}{2} \left( \sum_{n=1}^{\infty} \frac{1}{n} \right)$, which diverges ( the half factor occurs since $P$ consists of only the odd primes. Since the sequence of partial sums diverges anyway, it doesn't really matter).
The observation regarding the logarithm shows that the product exists if $s >2$, since $\ln(1+x) < x$ for $x >0$, so that the expression for $\ln(A)$ is less than $\sum_{p \in P} (p-1)^{1-s}$, which converges. However, for $1 < s < 2$, the product diverges, since, for a given $p$, the contribution to the product from $p$ is at least $1 + (p-1)^{1-s}$, (using the binomial theorem), so at least $\frac{p}{p-1}$. Hence the product is at least $\frac{1}{2} \left( \sum_{n=1}^{\infty} \frac{1}{n} \right)$, which diverges ( the half factor occurs since $P$ consists of only the odd primes. Since the sequence of partial sums diverges anyway, it doesn't really matter).