Following the comments from Did.
The statements aren't equivalent.
One direction holds, i.e. the first statement implies the second, as
$$\liminf\limits_{x\to\infty}\frac{\overline{F}(x)}{e^{-\lambda x}}>0\implies(\forall \mu>\lambda)\lim\limits_{x\to\infty}\frac{\overline{F}(x)}{e^{-\mu x}}=\lim\limits_{x\to\infty}\frac{\overline{F}(x)}{e^{-\lambda x}}e^{(\mu-\lambda) x}=\infty.$$
The other direction isn't true. A counterexample follows.
We'll define $$x_n=2^{4^n},\quad a_n=e^{-\sqrt{x_n}},\quad \frac1c=\sum\limits_{n=0}^{\infty}a_n,$$
and assume that $$P(X=x_n)=ca_n,\quad \forall n\in\mathbb{N}_0.$$
It follows that $x_{n+1}=x_n^4$ and $$\overline{F}(x_n)=\overline{F}(x_{n+1}-1)=c\sum\limits_{k=n+1}^{\infty}a_k \, \sim\,ca_{n+1}=ce^{-\sqrt{x_{n+1}}}=ce^{-x_n^2}.$$
Hence $$\overline{F}(x_n)\sim ce^{-x_n^2},\quad\overline{F}(x_n-1)\sim ce^{-\sqrt{x_{n}}}\sim ce^{-\sqrt{x_{n}-1}}.$$
Thus we have $$(\forall\lambda>0)\limsup\limits_{x\to\infty}\frac{\overline{F}(x)}{e^{-\lambda x}}\geq \limsup\limits_{n\to\infty}\frac{\overline{F}(x_n-1)}{e^{-\lambda (x_n-1)}}=\limsup\limits_{n\to\infty}\frac{ce^{-\sqrt{x_{n}-1}}}{e^{-\lambda (x_n-1)}}=\infty,$$
but $$\liminf\limits_{x\to\infty}\frac{\overline{F}(x)}{e^{-\lambda x}}\leq \liminf\limits_{n\to\infty}\frac{\overline{F}(x_n)}{e^{-\lambda x_n}}=\liminf\limits_{n\to\infty}\frac{ce^{-x_n^2}}{e^{-\lambda x_n}}=0.$$