Because the scale is too small in Mertens's theorem, and the prime number theorem as well as the Riemann hypothesis are hidden by the $O(1/\log{X})$ notation.

Indeed, the former amounts to strengthening this term to $o(1/\log{X})$; the latter - to $O(1/\sqrt{X})$.

[Incidentally, one could go for equivalent statements to a still smaller scale. Of course, an estimate like $\sum_{p < X} 1/p^2 = \mathrm{const} + O(1/X)$ does not say anything at all about the primes. But one could, if one wished, express the prime number theorem by elaborating on the $O(1/X)$ term. ]

To elaborate on this a bit, let me go to a slightly bigger scale where the prime number theorem begins to emerge outside of $o(1)$. This is also more natural; indeed, it was how Mertens's theorem was proved.

By partial summation, Mertens's estimate is equivalent to $\sum_{p < X} (\log{p})/(p-1) = \log{X} + O(1)$; or, if one prefers, $\sum_{n < X} \Lambda(n)/n = \log{X} + O(1)$. The prime number theorem however is the statement that the $O(1)$ term converges to a constant: $\sum_{n < X} \Lambda(n)/n = \log{X} - \gamma + o(1)$. Indeed, the related bound $\sum_{n < X} \Big( \frac{1}{n} - \frac{1}{X} \Big) \Lambda(n) = \log{X} - 1 - \gamma + o(1)$, another form of the prime number theorem, is what de la Vallee Poussin actually obtained in his original paper. Here $\gamma = 0.57\ldots$ is Euler's constant, but this is of no importance for us, see the next paragraph. Also the $\log{X}$ logarithmic pole corresponds to the pole of $\zeta(s)$ at $s=1$, whereas the $o(1)$ term expresses there being no zeros with $\mathrm{Re}(s) = 1$. The Riemann hypothesis is the correspondingly stronger bound $O(1/\sqrt{X})$ on the $o(1)$ oscillating term. At this scale, in contrast to $\psi(X) = X + O\big( \sqrt{X} (\log{X})^2 \big)$ or $\pi(X) = \mathrm{Li}(X) + O(\sqrt{X}\log{X})$, a logarithmic factor in addition to the square root is not required, as $\sum_{\rho} 1 / |\rho|^2 < \infty$ over the zeros.

Here finally is how to deduce the more usual form $\psi(X) \sim X$ of the prime number theorem from the refinement $S(X) := \sum_{p < X} \Lambda(n)/n = \log{X} -\gamma + o(1)$ of Mertens's theorem: Summing by parts, $\psi(X) = XS(X) - \int_1^X S(t) \, dt = X(\log{X} - \gamma + o(1)) - \big( \int_1^X \log{t} \, dt - \gamma X + o(X) \big) = X + o(X)$.

**Added** (December, 2017). I came upon an observation giving also a 'trivial' proof of the reverse elementary implication of the two purely qualitative forms, multiplicative and logarithmic, of the prime number theorem: $\psi(X) \sim X \Leftrightarrow S(X) = \log{X} - \gamma + o(1)$. The following seems to have been missed in the literature on elementary methods which, at this point, seem all to quote a somewhat more involved Tauberian theorem of Axer; cf. section 8.1.1 of Montgomery and Vaughan's book (*Multiplicative Number Theory: I*) or, for a more general setting, chapter 14 of Diamond and Zhang's recent book on *Beurling Generalized Numbers* (really this paper of theirs). The simpler argument below also extends easily to number fields, supplying a particularly easy proof of the 'elementary equivalence' of Landau's prime ideal theorem and number field sharp Mertens. Incidentally, as I happen to recall, this addresses a slightly curious point that had come up in the comments to this answer of Eric Naslund. Remembering also my answer here, I figured it may be worth to record the following observation as an addendum to it, sticking for simplicity to the rational case assumed in this question.

*A proof of $\psi(X) \sim X \Rightarrow S(X) = \log{X} - \gamma + o(1)$.* For simplicity, let me stick to $\mathbb{Q}$. The case of a number field $K$ has the same result with $\gamma$ generalized as the 'Euler-Kronecker invariant' $\gamma_K$.

The key is to observe that the formula
$$
X^{-1} \log{X!} = \sum_{n \leq X/T} \frac{\Lambda(n)}{n} + \sum_{m \leq T} \frac{1}{X} \Big( \psi\Big( \frac{X}{m} \Big) - \psi\Big( \frac{X}{T} \Big) \Big) + O(1/T)
$$
holds uniformly in the two parameters $X, T \geq 1$, with an absolute implied coefficient. It interpolates between Mertens's estimate (case $T = 1$) and Chebyshev's convolution formula $\log{X!} = \sum_m \psi(X/m)$ (case $T = \infty$). But the general formula also follows, after a moment of reflection, from Chebyshev's argument with the prime factorization of $X!$. Divide the moduli into the ranges $n \leq X/T$ and $n > X/T$. The total contribution of the latter are exactly accounted for by the second sum. For a small modulus $n \leq X/T$, the contribution via the prime factorization is $X^{-1} \lfloor X/n \rfloor \Lambda(n) = \frac{\Lambda(n)}{n} + O\Big(\frac{\Lambda(n)}{X}\Big)$, neglecting the fractional part. The $O(1/T)$ term then comes from summing these for $n \leq X/T$, and using Chebyshev's estimate $\sum_{n \leq Y} \Lambda(n) \ll Y$. (In the number field generalization, the latter estimates extend as lattice point counts.)

Now, by Stirling's asymptotic, the qualitative $\psi(X) \sim X \Rightarrow S(X) = \log{X} - \gamma + o(1)$ implication is immediate from the observed formula upon first letting $X \to \infty$ and then $T \to \infty$.

couldprove PNT. :) $\endgroup$1more comment