$\newcommand{\R}{\mathbb R}$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{1}\label{1}
\end{equation*}
where 
\begin{equation*}
	L^*(y):=\sup_{t\in\R}(ty-L(t)),\quad L(t):=\ln Ee^{tX}, \tag{2}\label{2}
\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{1}, $s(x)=0$ for all real $x\ge0$, so that the desired relation
\begin{equation*}
	s(x)\sim t(x)=\ln P(X>x) \tag{3}\label{3}
\end{equation*}
will not hold. 

So, without loss of generality (wlog)
\begin{equation}
	\text{$L(t_0)<\infty$ for some real $t_0>0$, } \tag{4}\label{4}
\end{equation}
and then, by \eqref{2}, $L^*(y)\ge t_0y-L(t_0)\to\infty$ as $y\to\infty$. Also, the function $L^*$ is convex, being the supremum of affine functions. So, by \eqref{1}, for all large enough real $x>0$ 
\begin{equation*}
	s(x)=-L^*(x)=\ln\inf_{t\in\R}Q(x,t),  \tag{1a}\label{1a}
\end{equation*}
where
\begin{equation*}
	Q(x,t):=e^{-tx}M(t),\quad M(t):=Ee^{tX}. \tag{5}\label{5}
\end{equation*}
The condition that $L(t_0)<\infty$ for some real $t_0>0$ implies 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)=T(x):=\ln\inf_{t\ge0}Q(x,t).   \tag{1b}\label{1b}
\end{equation*}

Let 
\begin{equation*}
	q(x):=P(X>x), 
\end{equation*}
so that 
\begin{equation*}
	t(x)=\ln q(x). \tag{5.5}\label{5.5}
\end{equation*}
By Markov's inequality, $q(x)\le Q(x,t)$ for all real $t\ge0$. So, by \eqref{1b},
\begin{equation*}
	s(x)\ge t(x) \tag{5.75}\label{5.75}
\end{equation*}
eventually (that is, for all large enough real $x>0$). 

It should be clear that, for \eqref{3} to hold, the tail function $q$ should vary regularly enough. (Indeed, the function $s$ is always concave; so, if, for instance, $t$ is a function decreasing to $-\infty$ in a neighborhood of $\infty$ so that $t$ is not asymptotically equivalent to **any** concave function, then, of course, \eqref{3} cannot hold.) On the other hand, it follows from \eqref{4} that $q$ cannot decrease near $\infty$ slower than all decreasing exponential functions. So, it seems reasonable to assume that the tail function $q$ is log concave in a neighborhood of $\infty$ (especially because its desirably "matching" function $e^s$ is log concave). It turns out that this is enough for \eqref{3}: 

>**Proposition 1:** If the tail function $q$ is log concave and nonzero in a neighborhood of $\infty$, then \eqref{3} holds. 

*Proof:* By \eqref{5}, 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{6}\label{6}
\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{7}\label{7}
\end{equation*}
\begin{equation*}
	I_2:=t\int_{x-A}^\infty \,dv\,e^{tv} e^{-g(v)}, \tag{8}\label{8}
\end{equation*}
$A$ is any real number, and $g:=-\ln q$, so that $g$ is (strictly) increasing, convex, $>0$, and $<\infty$ in a neighborhood of $\infty$ -- say on the interval 
\begin{equation*}
	[a,\infty)
\end{equation*}
for some real $a$. So, $g'>0$ on $[a,\infty)$, where $g'$ is (say) the right derivative of the convex function $g$. 
In what follows, by default $x\in[a,\infty)$. 

Take now any $c\in(0,1)$ and let  
\begin{equation*}
	A:=\frac{cg(x)}{g'(x)}, \quad t:=g'(x)-\frac1A=\frac{cg(x)-1}A, \tag{9}\label{9}
\end{equation*}
so that eventually $A>0$ and $t>0$. 
Note also that, by the convexity of $g$ on $[a,\infty)$, we have $(x-a)g'(x)\ge g(x)-g(a)$ (for $x\in[a,\infty)$), that is, $xg'(x)-g(x)\ge ag'(x)-g(a)$, so that  
\begin{equation*}
	x-A=(1-c)x+c\frac{xg'(x)-g(x)}{g'(x)}
	\ge(1-c)x+c\frac{ag'(x)-g(a)}{g'(x)}. \tag{10}\label{10}
\end{equation*}
Also, again by the convexity of $g$ on $[a,\infty)$, there exists $g(\infty):=\lim_{x\to\infty}g'(x)\in(0,\infty]$. So, by \eqref{10}, $x-A\to\infty$ and hence wlog $x-A\ge a$, which will be assumed by default in the sequel. 
Using the convexity of $g$ on $[a,\infty)$ again, we have 
\begin{equation}
	g(v)\ge h_x(v):=g(x)+g'(x)(v-x)
\end{equation}
for real $v\ge x-A$. So, by \eqref{8} and \eqref{9},
\begin{equation*}
	e^{-tx}I_2\le t\int_{x-A}^\infty \,dv\,e^{tv} e^{-h_x(v)} \\ 
	=e^{-g(x)+1}(cg(x)-1)=e^{-g(x)(1+o(1))}=q(x)^{1+o(1)} \tag{11}\label{11}
\end{equation*}
(as $x\to\infty$). 
Also, by \eqref{7} and \eqref{9},
\begin{equation*}
	e^{-tx}I_1\le e^{-tA}=e^{-cg(x)+1}=eq(x)^c.
\end{equation*}
Since $c\in(0,1)$ was arbitrary, we now get 
\begin{equation*}
	e^{-tx}I_1\le q(x)^{1+o(1)}. \tag{12}\label{12}
\end{equation*}
It follows from \eqref{1b}, \eqref{5}, \eqref{6}, \eqref{12}, \eqref{11}, and \eqref{5.5} that 
\begin{equation}
	s(x)\le\ln\big(q(x)^{1+o(1)}\big)\sim\ln q(x)=t(x). \tag{13}\label{13}
\end{equation}
Finally, \eqref{3} follows from \eqref{5.75} and \eqref{13}. $\quad\Box$ 

**Remark:** It seems that, following the lines of the above proof, one can show that it is necessary and sufficient for \eqref{3} that the function $t$ be asymptotically equivalent to some concave function. I am going to check this later, after getting some sleep. 

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