Heavy-tailed distributions
We will need to show that $(i)\implies (ii)$ and $(ii) \implies (i)$
Part 1: $(i) \implies (ii)$
Lets assume that $P(X>x)$ is a smoothly decreasing function bounded below by $0$.
Assume (i): $\limsup_{x\to\infty}\frac{P(X>x)}{e^{-\lambda x}} >0 \;\forall \lambda>0.$ Since $e^{-\lambda x},P(X>x)$ are positive, monotonic decreasing functions, $(i) \implies \forall \lambda>0,\{\exists c>0:P(X>x)\geq e^{-\lambda x}\;\forall x>c\}$
Now, we also know that $P(X>y|X>u)=\mathbf{1}_{<u}(y)+\frac{\mathbf{1}_{\geq u}(y)P(X>y)}{P(X>u)}$
Let $g(u):=E[X-u|X>u]$ where $g(u)\geq 0$ by definition of the conditional probability, then the above result (combined with the fact that $X\in [0,\infty)$, means that $g(u)=\int_0^{u}\mathbf{1}_{<u}(x)dx+\frac{\int_u^{\infty}P(X>x)dx}{P(X>u)}-u=\frac{\int_u^{\infty}P(X>x)dx}{P(X>u)}$
Combining this with our implication from $(i)$ gives us:
$\forall (\lambda>0),\exists c: \left\{g(u):=\frac{\int_u^{\infty}P(X>x)dx}{P(X>u)}\geq\frac{\int_u^{\infty} e^{-\lambda x}dx}{e^{-\lambda u}}=\int_u^{\infty} e^{-\lambda (x-u)} dx=\frac{1}{\lambda}\;\forall u>c\right\}$
However, $\lim_{\lambda \to 0} \frac{1}{\lambda} = \infty$; therefore, $g(u)$exceeds any bound, hence $\lim_{u\to\infty} g(u) = \infty$
Thus $(i)\implies (ii)$
Part 2: $(ii) \implies (i)$
Now, let $X$ be an arbitrary, positive continuous, integrable random variable, as you've specified in your post. Define the shifted conditional tail expectation $E[X-u|X>u]=E[X|X>u]-u=\frac{\int_u^{\infty}x f_X(x)dx}{P(X>u)}-u:=g(u)$
Assuming $(ii)$, we know that $g(u)$ must grow without bound, hence $g'(u)=O(\frac{1}{u^p}), p\leq1$. The most stringent value for $p$ is $p=1$, as it is on the threshold of convergence. Then we get:
$g'(u)=O(u^{-1})\implies g(u)=O(\ln(u))$. For concreteness, lets take $g(u)=\ln(u)$. This implies:
$\frac{\int_u^{\infty}x f_X(x)dx}{P(X>u)}-u = \ln(u) \implies \int_u^{\infty}x f_X(x)dx=(1-F_X(u))(\ln(u)+u)=\ln(u)+u -F_X(u)\ln(u)-uF_X(u)$.
Taking the derivative wrt $u$, we get:
$-uF_X'(u)=u^{-1}+1-\frac{F_X(u)}{u}-\ln(u)F_X'(u)-uF_X'(u)-F_X(u)$
Simplifying, we get:
$F_X'(u)=(1-F_X(u))\left[\frac{1}{u\ln(u)}+\frac{1}{\ln(u)}\right]$
Suppressing the fact that the distribution is for $X$ with argument $u$ and separating differentials, we get:
$\frac{dF}{1-F}=\frac{1+u}{u\ln(u)}du$
Integrating both sides we get:
$-\ln(1-F)+A=\ln(\ln(u))-\Re(\Gamma(0,-\ln(u)))+B$
Exponentiating both sides and simplifying, we get:
$\frac{e^A}{1-F}=\frac{e^B\ln(u)}{e^{\Re(\Gamma(0,-\ln(u))})}$
A little more algebra to isolate $(1-F)$, and we get (after combining the two unknown constants of integration):
$\frac{Ce^{\Re(\Gamma(0,-\ln(u)))}}{\ln(u)}=1-F=P(X>u)$
Thus $F(x) = 1-\frac{Ce^{\Re(\Gamma(0,-\ln(u)))}}{\ln(u)}$. If we set $C=1$, this function is a valid distribution function for a random variable defined on $[1.72719,\infty)$. Its plot is shown below:
Note that it crosses the $x-$axis at $x\approx 1.72719$, so the random variable's distribution function would be $P(X<x)= \mathbf{1}_{>1.72719}(x)F(x)$
Unfortunately, $P(X>x)=1-\mathbf{1}_{>1.72719}(x)F(x)$ shrinks faster than an exponential, hence it will not satisfy $(i)$. Thus, $(ii) \nRightarrow (i)$
$\square$
Therefore, it appears that $(i)$ is a stronger definition than $(ii)$, in the sense that it implies $(ii)$, which is a fact that makes sense given a distribution is heavy tailed (i.e., $(ii)$ should apply to all heavy-tailed distribution). However, $(ii)$ also applies to distributions that do not satisfy $(i)$, which is also a reasonable criteria for a heavy-tailed distribution.
The definition I've seen most often is $(i)$, and I think the above demonstration shows why that's the case. It's a difference between what properties are necessary vs sufficient. $(i)$ is sufficient while $(ii)$ appears merely necessary.
Please let me know if any of the above steps do not make sense, or, as per @Did, there are gaps or oversights...I'm not a professional "proof-writer", but this problem seemed very interesting so I wanted to give it a go, especially since you haven't received other answers.