Let $s_0>0$. The right statement is that the following are equivalent:
The sum $\sum_{n=1} \tfrac{\mu(n)}{n^s}$ converges for $s>s_0$.
$\zeta(s)$ has no zeroes with real part $>s_0$.
$1/\zeta(s)$ has an analytic continuation to the half space $\mathrm{Re}(s) > s_0$
$\sum_{n=1}^N \mu(n) = O(N^{s_0+\epsilon})$ for all $\epsilon >0$.
As you point out, $1/\zeta(s)$ certainly has an analytic continuation to the real interval $(0,\infty)$, since $\zeta$ doesn't vanish on the positive reals.