The standard proof of the Prime Number Theorem requires taking into consideration that there are no zeroes of the Riemann Zeta function in which the real part equals one. But consider the following argument:
The probability that a number less than X is prime $\pi(X)/X$ is approximately $\Pi_{p<X} (1-1/p)$ (this is for the same reason that the Sieve of Eratosthenes works), which is approximately $1/\Sigma_{n=1}^X 1/n$, which is approximately $1/\log X$. Hence, $\pi(X)$ is approximately $X/\log X$.
This doesn't use complex numbers, but it gives a good reason to believe the Prime Number Theorem. Why do complex numbers (which seem to come from nowhere) make this argument rigorous?
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.