Here is a heuristic which I find useful for ruling out easy proofs of the PNT. Consider a set of positive integers $P$ with the following properties: Between $2^{2k-1}$ and $2^{2k}$, there are roughly $a \frac{2^{2k-1}}{(2k-1) \log 2}$ elements of $P$ and, between $2^{2k}$ and $2^{2k+1}$, there are roughly $b \frac{2^{2k}}{(2k) \log 2}$ elements of $P$.
Now, if $P$ is the primes, then $a=b=1$. Suppose instead that $a=1 + c$ and $b=1 - c$, for some small constant $c$.
Then $\prod_{p \in P} (1-p^{-s})^{-1}$ has a simple pole at $s=1$, with residue $1$. The sum $\sum_{p \in P,\ p \leq N} 1/p$ grows like $\log \log N$. And, regarding your specific question, $\prod_{p \in P,\ p \leq N} (1-1/p) \approx 1/\log N$.1 So these properties can't distinguish $P$ from the set of primes.
However, the PNT does not hold for $P$. Let $\pi_P(N)$ be the number of elements of $P$ which are $\leq N$. Then, if $N=2^{2k}$, then
$$\frac{\pi_P(N)}{N/\log N} = (2k) \left( a \frac{2^{2k-1}}{2k-1} + b \frac{2^{2k-2}}{2k-2} + a \frac{2^{2k-3}}{2k-3} + \cdots \right)$$
$$\approx a \left( \frac{1}{2} + \frac{1}{8} + \cdots \right) + b \left( \frac{1}{4} + \frac{1}{16} + \cdots \right) = (2/3) a + (1/3) b. $$
Similarly, if $N=2^{2k+1}$, then
$$\frac{\pi_P(N)}{N/\log N} \approx (2/3) b + (1/3) a.$$
So the ratio of $\pi_P(N)/(N/\log N)$ does not approach a well defined ratio. Any proof of the PNT must use facts about the primes which distinguish them from $P$.
1 There is also a second issue here. It turns out that
$$\prod_{p\ \mbox{Prime},\ p \leq N} \left(1-\frac{1}{p} \right) \sim \frac{e^{- \gamma}}{\sum_{n \leq N} 1/n},\ \mbox{not}\ \sim \frac{1}{\sum_{n \leq N} 1/n}.$$
So you would have to explain why that $e^{- \gamma}$ disappears.