I'm not an expert in this area, but this may be a start.
Rather than $\prod_{p\lt n}$, you can use $\prod_{p\le \sqrt n}$.
$\log\log \sqrt n + \gamma \lt \log\log \sqrt n +\log 2 = \log\log n $
That gets you a little closer, since now you are off by $\log 2 - \gamma \approx 0.116$.
The heuristic probability that $n$ is prime is not
$$\prod_{p\lt n} (1-Pr(p|n))$$
It is the product of probabilities
$$\prod_{p\lt n} (1-Pr(p|n \text{ given no smaller prime divides } n))$$
For $p$ small, the term you get may be close to $(1-1/p)$, but I that's not the case for $p$ large.
For $\sqrt n \lt p$, the term corresponding to $p$ in the product is just $1$.
For $\sqrt[3]n \lt p \le \sqrt n$, if $p$ is the smallest prime dividing $n$, then $n/p$ must be prime, too. Perhaps that means that by strong induction, we should discount these terms by the probability $n/p$ is prime, about $1/\log \frac np$, so that those terms in the product are $(1-1/(p \log \frac np))$.
It looks like you get some sums/integrals if you try to extend this to more terms, and I don't know whether you can expect to get the desired accuracy at the end.
Here is a heuristic that I plan to make into a blog post some day. Suppose that there were only finitely many primes with first digit $9$. Is your estimate good enough to see that?
To be more precise, suppose that there were no primes between $9 \times 10^k$ and $10^{k+1}$ for all sufficiently large $k$. And suppose that the number of primes between $a$ and $b$, for $10^k \leq a < b \leq 9 \times 10^k$ is $\approx \frac{\log 10}{\log 9} \int_{a}^b dt/\log t$ (when $a$ is not too close to $b$). We'll see later where the fraction $\log 10/\log 9$ comes from.
The first thing to note is that this would violate the prime number theorem. In this scenario, we would have $\pi(9 \times 10^k) = \pi(10 \times 10^k)$ for $k$ large. But the prime number theorem says that
$$\pi(10\times 10^k) - \pi(9 \times 10^k) \sim \frac{10 \times 10^k}{k \log 10 + \log {10}} - \frac{9 \times 10^k}{k \log 10 + \log {9}} \sim \frac{10^k}{k \log 10} \to \infty.$$
So proving the prime number theorem involves disproving this ridiculous scenario.
Now, let's see that the scenario is consistent with $\sum_{p \leq N} 1/p = \log \log N + M + O(1/\log N)$. The sum over the primes between $10^k$ and $10^{k+1}$ would be roughly
$$\frac{\log 10}{\log 9} \int_{10^k}^{9 \times 10^k} \frac{dt}{t \log t} = \frac{\log 10}{\log 9} \left( \log \log (9 \times 10^k) - \log \log 10^k \right)$$
$$=\frac{\log 10}{\log 9} \left( \log( k \log 10+\log 9) - \log (k \log 10) \right) = \frac{\log 10}{\log 9} \left( \log \left( 1+\frac{\log 9}{k \log 10} \right) \right) $$
$$ = \frac{\log 10}{\log 9} \frac{\log 9}{k \log 10} + O(1/k^2)= \frac{1}{k} + O(1/k^2)$$
So
$$\sum_{p \leq 10^{n+1}} \frac{1}{p} = \sum_{j=1}^n \left( \frac{1}{j} + O(1/j^2) \right)=$$
$$\log n + B + O(1/n) = \log \log 10^{n+1} + C + O(1/\log 10^{n+1}).$$
Very important exercise left for you: If you redo this computation for $\sum_{p \leq 9 \times 10^k} 1/p$, you get $\log \log (9 \times 10^k) + C + O(1/\log(9 \times 10^k))$ for the same constant $C$. The point is that $\log \log 10^{k+1} - \log \log (9 \times 10^k) = O(1/\log 10^k)$, so this estitmate is consistent with $\sum_{p \leq N} 1/p$ not growing at all between $9 \times 10^k$ and $10 \times 10^k$.
This trick is useful for refuting other simple approaches to the PNT. For example, the "primes hate to start with $9$ scenario" is also consistent with $\sum \log p/p^s = 1/(s-1) + O(1)$ as $s \to 1^{+}$, so that is also not enough to prove PNT.
Best Answer
Well, one can always say that the PNT is equivalent to $$\sum_{p \leq x}\frac{1}{p} = \log \log x + M + o\left(\frac{1}{\log x}\right),\tag{$\ast$}$$ because both results are true (with better error terms). This is of course not what is meant by the Wikipedia page. Instead, the idea is that the equivalence PNT$\,\Leftrightarrow(\ast)$ can be established in a simpler way than either PNT or $(\ast)$. On the other hand, "simpler" is a subjective word, e.g. I usually find Tauberian arguments tricky.
At any rate, the first three exercises in Section 8.1.1 of Montgomery-Vaughan: Multiplicative number theory I address this question. For example, the PNT easily implies the relation $\psi(x)\sim x$ (logarithmic weights!), which then implies rather nontrivially (using Theorem 8.1 = Axer's theorem) that $$\sum_{p\leq x}\frac{\log p}{p}=\log x+C+o(1)$$ for some constant $C$. From here, $(\ast)$ follows easily by partial summation.