$\newcommand\R{\mathbb R}$You are not saying for what values of $s$ the convergence takes place. Let us assume that it holds for all $s\in\R_+:=[0,\infty)$. That is, let us assume that 
\begin{equation*}
	L_n(s)\to L(s) \tag{1}\label{1}
\end{equation*}
(as $n\to\infty$) for each $s\in\R_+$, where 
\begin{equation*}
	L_n(s):=\int_{\R_+}e^{-sx}\mu_n(dx),\quad L(s):=\int_{\R_+}e^{-sx}\mu(dx). 
\end{equation*} 

From the context, $\mu$ and the $\mu_n$'s are finite measures over $\R_+$. You are not saying whether these measures are probability measures. 

If they are, then, by (say) [Theorem 5.22][1], condition \eqref{1} implies the weak convergence of $\mu_n$ to $\mu$; that is, we have 
\begin{equation*}
	\int_{\R_+}f(x)\mu_n(dx)\to\int_{\R_+}f(x)\mu(dx) \tag{2}\label{2}
\end{equation*}
for all bounded continuous functions $f$ on $\R_+$ -- which of course implies the vague convergence, the latter being the convergence \eqref{2} for all continuous functions $f$ on $\R_+$ such that $f(x)\to0$ as $x\to\infty$. 

---

Suppose now that $\mu$ and the $\mu_n$'s are any finite measures over $\R_+$ (not necessarily probability measures) such that \eqref{1} holds for each $s\in\R_+$. In particular, we have $\mu_n(\R_+)=L_n(0)\to L(0)=\mu(\R_+)$. 

Consider first the case when $\mu(\R_+)=0$. Take any bounded continuous function $f$ on $\R_+$, so that $|f|\le c$ for some real $c$. Then 
$$\Big|\int_{\R_+}f(x)\mu_n(dx)\Big|\le c\mu_n(\R_+)\to c\mu(\R_+)=0.$$ 
So, $\int_{\R_+}f(x)\mu_n(dx)\to0=\int_{\R_+}f(x)\mu(dx)$. Thus, $\mu_n$ converges to $\mu$ weakly and hence vaguely. 

Finally, consider the case when $\mu(\R_+)>0$. Then eventually (that is, for all large enough $n$) we have $\mu_n(\R_+)>0$. So, (for such $n$) we may introduce the probability measures 
\begin{equation*}
	\nu_n:=\frac{\mu_n}{\mu_n(\R_+)},\quad \nu:=\frac{\mu}{\mu(\R_+)}. 
\end{equation*}
Since $\mu_n(\R_+)\to\mu(\R_+)>0$, we see that the Laplace transform of $\nu_n$ will converge to the Laplace transforms of the $\nu$. So, by what was said in the case when $\mu$ and the $\mu_n$'s are probability measures, $\nu_n$ will converge to $\nu$ weakly. Again, since $\mu_n(\R_+)\to\mu(\R_+)>0$, it will follow that in this case as well $\mu_n$ converges to $\mu$ weakly and hence vaguely. $\quad\Box$ 

[1]: https://www.google.com/books/edition/Foundations_of_Modern_Probability/L6fhXh13OyMC?hl=en&gbpv=1&bsq=Theorem%205.22