$\newcommand{\thh}{\theta}$In general, you cannot get the vague convergence here. E.g., suppose that $\mu=0$ and $\mu_n(dx)=x^2\,1(n<x<n+1)\,dx$. Then for each real $s>0$
\begin{equation*}
0\le\int_0^\infty e^{-sx}\mu_n(dx)=\int_n^{n+1} e^{-sx}x^2\,dx
\le e^{-sn}(n+1)^2\to0
\end{equation*}
(as $n\to\infty$), so that
\begin{equation*}
\int_0^\infty e^{-sx}\mu_n(dx)\to\int_0^\infty e^{-sx}\mu(dx).
\end{equation*}
However, letting $f(x):=\frac1{1+x}$ for real $x\ge0$, we have $f\in C_0([0,\infty))$ and
\begin{equation*}
\int_0^\infty f(x)\mu_n(dx)=\int_n^{n+1} f(x)x^2\,dx
\ge \frac1{1+n+1}n^2\to\infty,
\end{equation*}
so that
\begin{equation*}
\int_0^\infty f(x)\mu_n(dx)\to\infty\ne0=\int_0^\infty f(x)\mu(dx).
\end{equation*}
However, assuming that the measure $\mu$ is finite, you can get the following:
\begin{equation*}
\int_0^\infty f(x)\mu_n(dx)\to\int_0^\infty f(x)\mu(dx) \tag{$*$}\label{1}
\end{equation*}
for all $f\in C_c([0,\infty))$, that is, for all continuous functions $f$ on $[0,\infty)$ with a compact support.
Indeed, take any such $f$, so that $f=0$ on $[a,\infty)$ for some real $a>0$. Consider first the case when $f\ge0$.
Take any real $h>0$ and let
\begin{equation*}
\nu_{n,h}(dx):=e^{-hx}\mu_n(dx),\quad \nu_h(dx):=e^{-hx}\mu(dx).
\end{equation*}
Then for any real $s\ge0$
\begin{equation*}
\int_0^\infty e^{-sx}\nu_{n,h}(dx)=\int_0^\infty e^{-(s+h)x}\mu_n(dx)
\to\int_0^\infty e^{-(s+h)x}\mu(dx)=\int_0^\infty e^{-sx}\nu_h(dx).
\end{equation*}
So, by the previous answer, $\nu_{n,h}\to\nu_h$ weakly. So,
\begin{equation*}
\int_0^\infty f(x)e^{-hx}\mu_n(dx)=\int_0^\infty f(x)\nu_{n,h}(dx)
\to\int_0^\infty f(x)\nu_h(dx)=\int_0^\infty f(x)e^{-hx}\mu(dx)
\end{equation*}
for each real $h>0$.
Next,
\begin{equation*}
\int_0^\infty f(x)\mu_n(dx)\ge\int_0^\infty f(x)e^{-hx}\mu_n(dx)
=\int_0^a f(x)e^{-hx}\mu_n(dx) \\
\ge e^{-ha}\int_0^a f(x)\mu_n(dx)
=e^{-ha}\int_0^\infty f(x)\mu_n(dx),
\end{equation*}
so that
\begin{equation*}
\int_0^\infty f(x)\mu_n(dx)=e^{\thh_{n,h,f}ha}\int_0^\infty f(x)e^{-hx}\mu_n(dx)
\end{equation*}
for some $\thh_{n,h,f}\in[0,1]$ depending on $n,h,f$. Similarly,
\begin{equation*}
\int_0^\infty f(x)\mu(dx)=e^{\thh_{h,f}ha}\int_0^\infty f(x)e^{-hx}\mu(dx)
\end{equation*}
for some $\thh_{h,f}\in[0,1]$ depending on $h,f$. So, for each real $h>0$,
\begin{equation*}
\begin{aligned}
\limsup_n\int_0^\infty f(x)\mu_n(dx)&\le e^{ha}\limsup_n\int_0^\infty f(x)e^{-hx}\mu_n(dx) \\
& =e^{ha}\int_0^\infty f(x)e^{-hx}\mu(dx) \\
& \le e^{ha}\int_0^\infty f(x)\mu(dx) \\
& \underset{h\downarrow0}\longrightarrow
\int_0^\infty f(x)\mu(dx)
\end{aligned}
\end{equation*}
and
\begin{equation*}
\begin{aligned}
\liminf_n\int_0^\infty f(x)\mu_n(dx)&\ge\liminf_n\int_0^\infty f(x)e^{-hx}\mu_n(dx) \\
&=\int_0^\infty f(x)e^{-hx}\mu(dx) \\
&\ge e^{-ha}\int_0^\infty f(x)\mu(dx) \\
& \underset{h\downarrow0}\longrightarrow
\int_0^\infty f(x)\mu(dx).
\end{aligned}
\end{equation*}
So, $\limsup_n\int_0^\infty f(x)\mu_n(dx)=\liminf_n\int_0^\infty f(x)\mu_n(dx)=\int_0^\infty f(x)\mu(dx)$ and hence \eqref{1} holds
for any nonnegative $f\in C_c([0,\infty))$.
Finally, writing $f=f_+-f_-$, where $f_\pm:=\max(0,\pm f)$, we get \eqref{1} for all $f\in C_c([0,\infty))$. $\quad\Box$
