You ask:

Denote by $\mu$ the Mobius function. It is known that for every integer $k>1$, the number $\sum_{n=1}^{\infty} \frac{\mu(n)}{n^k}$ can be interpreted as the probability that a randomly chosen integer is $k$-free.

Letting $k\rightarrow 1^+$, why shouldn't this entail the Prime Number Theorem in the form

$$\sum_{n=1}^{\infty} \frac{\mu(n)}{n}=0,$$

since the probability that an integer is ``$1$-free'' is zero ?

As pointed out by the users @wojowu and @PeterHumphries,
it is true that the PNT is equivalent to

$$\lim_{x \to \infty} \sum_{n\leq x} \frac{\mu(n)}{n}=0,$$
and it is relatively easy to prove that

$$\lim_{s\rightarrow 1^+} \sum_{n=1}^{\infty} \frac{\mu(n)}{n^s}=0.$$
The real difficulty lies in proving that

$$\lim_{x\rightarrow \infty} \sum_{n\leq x} \frac{\mu(n)}{n}=
\lim_{s\rightarrow 1^+} \sum_{n=1}^{\infty} \frac{\mu(n)}{n^s},$$
which is highly nontrivial and requires intricate arguments.

In particular, as pointed out by @TerryTao in the comments:

if $t\neq 0$ is real, then

$$ \lim_{s\rightarrow 1^+} \sum_{n=1}^{\infty} \frac{n^{it}}{n^s},$$

can be shown to converge to a finite value, whereas
$$\lim_{x\rightarrow \infty} \sum_{n\leq x} \frac{n^{it}}{n}$$

is undefined. So at a bare minimum one has to somehow stop $\mu(n)$ from "pretending" to be like $n^{it}$. This turns out to be basically equivalent to preventing $\zeta(s)$ from having a zero at $1+it$, and actually showing this doesn't occur is at the very heart of proving the PNT.

istrue that the PNT is equivalent to $\sum_{n \leq x} \frac{\mu(n)}{n} = o(1)$. It is also relatively easy to prove that $\lim_{s \searrow 1} \sum_{n = 1}^{\infty} \frac{\mu(n)}{n^s} = 0$. The hard part is proving that $\lim_{s \searrow 1} \sum_{n = 1}^{\infty} \frac{\mu(n)}{n^s} = \lim_{x \to \infty} \sum_{n \leq x} \frac{\mu(n)}{n}$. This is highly nontrivial! $\endgroup$ – Peter Humphries Apr 20 '19 at 21:48