$\newcommand{\R}{\mathbb R}$
>**Proposition 1:** For 
\begin{equation*}
	s(x)\sim T(x)=\ln P(X>x) \tag{00}\label{00}
\end{equation*}
to hold (as $x\to\infty$), it is necessary and sufficient that 
\begin{equation*}
	T(x)\sim T_0(x) \tag{10}\label{10}
\end{equation*}
for some real-valued concave function $T_0$. 

The relation $a\sim b$ is understood here as $a/b\to1$. In particular, $a\sim b$ implies that eventually the values of $|b|$ are not in the set $\{0,\infty\}$. 

*Proof of Proposition 1:* 

**Part 1: Necessity:** Assume that \eqref{00} holds; in particular, this implies that $T(x)\ne-\infty$ eventually (that is, for all large enough $x>0$); it is clear that $T(x)\ne0$ eventually. We have to show that then \eqref{10} holds as well. 

By Cram´er's large deviation principle (see e.g. [Corollary 2.2.19][1]), for all real $x$ 
\begin{equation*}
	s(x)=-\inf_{y\in[x,\infty)}L^*(y), \tag{20}\label{20}
\end{equation*}
where 
\begin{equation*}
	L^*(y):=\sup_{t\in\R}(ty-L(t)),\quad L(t):=\ln Ee^{tX}, \tag{30}\label{30}
\end{equation*}
$X:=X_1$. 

If $L(t)=\infty$ for all real $t>0$, then $L^*(y)=0$ for all real $y\ge0$, and hence, by \eqref{20}, $s(x)=0$ for all real $x\ge0$, so that \eqref{00} will not hold. 

So, without loss of generality (wlog)
\begin{equation}
	\text{$L(t_0)<\infty$ for some real $t_0>0$, } \tag{40}\label{40}
\end{equation}
and then, by \eqref{30}, $L^*(y)\ge t_0y-L(t_0)\to\infty$ as $y\to\infty$. Also, the function $L^*$ is convex, being the supremum of affine functions. So, by \eqref{20}, for all large enough real $x>0$ 
\begin{equation*}
	s(x)=-L^*(x)=\ln\inf_{t\in\R}Q(x,t),  \tag{20a}\label{20a}
\end{equation*}
where
\begin{equation*}
	Q(x,t):=e^{-tx}M(t),\quad M(t):=Ee^{tX}. \tag{50}\label{50}
\end{equation*}
Letting $T_0:=-L$ in a neighborhood of $\infty$ and looking at the first equality in \eqref{20a}, we see that \eqref{00} does imply \eqref{10}, for such a real-valued concave function $T_0$, which completes the necessity part of the proof. 

**Part 2: Sufficiency:** Assume that \eqref{10} holds for some real-valued concave function $T_0$. We have to show that then \eqref{00} holds as well. 

Since $T(x)\to-\infty$ (as $x\to\infty$), \eqref{10} and the concavity of $T_0$ imply that for some $\tau\in[-\infty,0)$ we have $T_0(x)/x\to\tau$ and $T(x)/x\to\tau$. So, \eqref{40} and \eqref{20a} hold. 

It also follows that there exists $\mu:=EX\in[-\infty,\infty)$. So, by Jensen's inequality, for $x>\mu$ and $t<0$ we have $Q(x,t)\ge e^{t(\mu-x)}>1=Q(x,0)$. So, for all large enough real $x$ 
\begin{equation*}
	s(x)=-L^*(x)=\ln\inf_{t\ge0}Q(x,t).   \tag{20b}\label{20b}
\end{equation*}

Let 
\begin{equation*}
	q(x):=P(X>x), 
\end{equation*}
so that 
\begin{equation*}
	T(x)=\ln q(x). \tag{60}\label{60}
\end{equation*}
By Markov's inequality, $q(x)\le Q(x,t)$ for all real $t\ge0$. So, by \eqref{20b},
\begin{equation*}
	s(x)\ge T(x) \tag{70}\label{70}
\end{equation*}
eventually (that is, for all large enough real $x>0$). 

By \eqref{50}, for real $t>0$,  
\begin{equation*}
\begin{aligned}
	M(t)=Ee^{tX}&=-\int_{-\infty}^\infty dq(u)e^{tu} \\ 
	&=-\int_{-\infty}^\infty dq(u)\int_{-\infty}^u t\,dv\,e^{tv}	\\ 
	&=-\int_{-\infty}^\infty t\,dv\,e^{tv}\int_v^\infty dq(u)	\\ 
	&=t\int_{-\infty}^\infty \,dv\,e^{tv} q(v) \\ 
	&=I_1+I_2,
\end{aligned}
\tag{80}\label{80}
\end{equation*}
where 
\begin{equation*}
	I_1:=t\int_{-\infty}^{x-A} \,dv\,e^{tv} q(v)
	\le t\int_{-\infty}^{x-A} \,dv\,e^{tv}=e^{t(x-A)}, \tag{90}\label{90}
\end{equation*}
\begin{equation*}
	I_2:=t\int_{x-A}^\infty \,dv\,e^{tv} e^{-g(v)}, \tag{100}\label{100}
\end{equation*}
$A$ is a real number (to be specified later), and 
$$g(v):=-\ln q(v)=-T(v).$$ 
Let 
$$g_0(v):=-T_0(v).$$  

The function $g_0$ is convex, (strictly) increasing, $>0$, and $<\infty$ in a neighborhood of $\infty$ -- say on the interval 
\begin{equation*}
	[a,\infty)
\end{equation*}
for some real $a$. So, $g_0'>0$ on $[a,\infty)$, where $g_0'$ is (say) the right derivative of the convex function $g_0$. 
In what follows, by default $x\in[a,\infty)$. 

Take now any $c\in(0,1)$ and let  
\begin{equation*}
	A:=A(x):=\frac{cg_0(x)}{g_0'(x)}, \tag{110}\label{110}
\end{equation*}
so that eventually $A>0$. 
Note also that, by the convexity of $g_0$ on $[a,\infty)$, we have $(x-a)g_0'(x)\ge g_0(x)-g_0(a)$ (for $x\in[a,\infty)$), that is, $xg_0'(x)-g_0(x)\ge ag_0'(x)-g_0(a)$, so that  
\begin{equation*}
	x-A=(1-c)x+c\frac{xg_0'(x)-g_0(x)}{g_0'(x)} \\ 
	\ge(1-c)x+c\frac{ag_0'(x)-g_0(a)}{g_0'(x)}. \tag{120}\label{120}
\end{equation*}
Also, again by the convexity of $g_0$ on $[a,\infty)$, there exists $g_0(\infty):=\lim_{x\to\infty}g_0'(x)\in(0,\infty]$. So, by \eqref{120}, $x-A\to\infty$ and hence wlog $x-A\ge a$, which will be assumed by default in the sequel. 

Let now 
\begin{equation*}
	k(x):=\inf_{v\in[x-A,\infty)}\frac{g(v)}{g_0(v)}, \tag{130}\label{130}
\end{equation*}
so that, in view of \eqref{10} and because $x-A\to\infty$, we have 
\begin{equation*}
	k(x)\to1. \tag{140}\label{140}
\end{equation*}

Now comes the **crucial point**, which is choosing the value of $t$ as follows: 
\begin{equation*}
	t:=t(x):=k(x)g_0'(x)-\frac1A=\frac{ck(x)g_0(x)-1}A. \tag{150}\label{150}
\end{equation*}

Using the convexity of $g_0$ on $[a,\infty)$ again, we have 
\begin{equation*}
	g_0(v)\ge h_x(v):=g_0(x)+g_0'(x)(v-x)
\end{equation*}
for real $v\ge x-A$. So,   
by \eqref{100}, \eqref{130}, \eqref{150}, \eqref{140}, and \eqref{10},
\begin{equation*}
	e^{-tx}I_2\le \int_{x-A}^\infty \,dv\,e^{tv} e^{-k(x)h_x(v)} \\ 
	=e^{-k(x)g_0(x)+1}(ck(x)g_0(x)-1)=e^{-k(x)g_0(x)(1+o(1))}=q(x)^{1+o(1)} \tag{160}\label{160}
\end{equation*}
(as $x\to\infty$). 
Also, by \eqref{90} and \eqref{150},
\begin{equation*}
	e^{-tx}I_1\le e^{-tA}=e^{-ck(x)g_0(x)+1}=eq(x)^{ck(x)}.
\end{equation*}
In view of \eqref{140} and because $c\in(0,1)$ was arbitrary, we now get 
\begin{equation*}
	e^{-tx}I_1\le q(x)^{1+o(1)}. \tag{170}\label{170}
\end{equation*}
It follows from \eqref{20b}, \eqref{50}, \eqref{80}, \eqref{170}, \eqref{160}, and \eqref{60} that 
\begin{equation*}
	s(x)\le\ln\big(q(x)^{1+o(1)}\big)\sim\ln q(x)=T(x). \tag{180}\label{180}
\end{equation*}
Finally, \eqref{00} follows from \eqref{70} and \eqref{180}. 

This completes the sufficiency part of the proof as well. 
$\quad\Box$ 

[1]: https://link.springer.com/book/10.1007/978-3-642-03311-7