In another answer here,
GH from MO shows PNT implies $\sum_{n \leq x} 1/p = \log x + C + o(1)$ for some constant $C$ (a hard step), and that implies $\sum_{p \leq x} 1/p = \log\log x + M + o(1/\log x)$ for some constant $M$ by partial summation. Here is a converse argument, from the estimate on $\sum_{p\leq x}1/p$ with error term $o(1/\log x)$ to PNT, thus establishing that there is an equivalence
$$
{\rm PNT} \Longleftrightarrow \sum_{p \leq x} \frac{1}{p} = \log\log x + M + o\left(\frac{1}{\log x}\right)
$$
for some constant $M$.

Set $A(x) = \sum_{p \leq x} 1/p$, so we assume $A(x) = \log\log x + M + o(1/\log x)$ for some $M$, as $x \to \infty$. Then
$$
\pi(x) = \sum_{p \leq x} 1 = \sum_{p \leq x} \frac{1}{p}p = \sum_{n \leq x} a_n n
$$
where $a_n = 1/p$ when $n = p$ is prime and $a_n = 0$ otherwise. Then $A(x) = \sum_{n \leq x} a_n$, so by partial summation
$$
\sum_{n \leq x} a_n n = A(x)x - \int_2^x A(y)\,dy
$$
for $x \geq 2$. Since $A(x) = \log\log x + M + o(1/\log x)$,
$$
A(x)x = x\log\log x + Mx + o\left(\frac{x}{\log x}\right)
$$
and
$$
\int_2^x A(y)\,dy = \int_2^x \log\log y\,dy + M(x-2) + \int_2^x o\left(\frac{1}{\log y}\right)\,dy.
$$
Using integration by parts,
$$
\int_2^x \log\log y\,dy = x\log\log x - 2\log\log 2 - \int_2^x\frac{dy}{\log y}.
$$
Putting these formulas into the expression for $\pi(x)$ as $\sum_{n \leq x} a_n n$, we get
$$
\pi(x) = \int_2^x \frac{dy}{\log y} + 2M + 2\log\log 2 + o\left(\frac{x}{\log x}\right) - \int_2^x o\left(\frac{1}{\log y}\right)\,dy.
$$
The constant $2M + 2\log\log 2$ can be absorbed into the $o(x/\log x)$ term.

Lastly, since $\int_2^x o(1/\log y)\,dy = o(\int_2^x dy/\log y)$ as $x \to \infty$, we have
$$
\pi(x) = \int_2^x \frac{dy}{\log y} + o\left(\frac{x}{\log x}\right),
$$
which is PNT.

PS. The Wikipedia page you mention no longer refers to an equivalence between PNT and the estimate on $\sum_{p \leq x} 1/p$ with error term $o(1/\log x)$, but only a one-way implication from PNT to the estimate on $\sum_{p \leq x} 1/p$: the View History tab of that page shows an edit was made to it on Jan. 10, 2023 by RStanley31 (gee, who could that be... :)) with the comment "Result of equivalence is not found within the literature." Perhaps that page should be reverted to the original wording with a link to this page.