$\newcommand{\R}{\mathbb R}$By Cramér's large deviation principle (see e.g. Corollary 2.2.19), 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. 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 $\infty$. It turns 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}. $\quad\Box$