$\newcommand\ep\varepsilon$

So the question can be restated as follows: if the Laplace transforms converge pointwise, do the CDFs also converge pointwise?

The answer is yes.

Indeed, let $P_n$ ($n=1,2,\ldots$) and $P$ be any probability distributions supported on a finite interval $[a,b]$ such that $LP_n\to LP$ pointwise on an interval $[0,T]$ for some real $T>0$, where $LQ$ is the Laplace transform of a probability distribution $Q$, so that $$(LQ)(t):=\int_{[a,b]}e^{-tx}Q(dx)$$ for real $t$.

Take any subsequence $(P_{n_k})$ of the sequence $(P_n)$ that converges weakly to a probability distribution $D$. Since $e^{-tx}$ is continuous in $x$, it follows that $(LP_{n_k})(t)\to (LD)(t)$ for all real $t$. So, $$LP=LD \text{ on } [0,T]. \tag{10}\label{10}$$

Since the distribution $P$ is supported on a finite interval, it is determined by its moments $m_k(P)$ ($k=1,2,\ldots$) -- see e.g. Theorem 30.1 in Billingsley. But $$m_k(P)=(-1)^k (LP)^{(k)}(0)$$ for $k=0,1,\dots$, so that the $m_k(P)$'s are in turn determined by the values of $LP$ on $[0,T]$. So, $P$ is determined by the values of $LP$ on $[0,T]$. So, by \eqref{10}, $D=P$. So, any weakly convergent subsequence of the sequence $(P_n)$ weakly converges to $P$.

Also, since all the $P_n$'s are supported on a finite interval, the sequence $(P_n)$ is tight.

So, by the Corollary to Theorem 29.3 on p. 381 in Billingsley's book, $(P_n)$ weakly converges to $P$.

That is, the c.d.f. $C_n$ of $P_n$ converges to the c.d.f. $C$ of $P$ at every point of continuity of $C$.

Finally, if $C$ is absolutely continuous (or just continuous), then $C_n\to C$ pointwise. $\quad\Box$

$\newcommand\ep\varepsilon$

So the question can be restated as follows: if the Laplace transforms converge pointwise, do the CDFs also converge pointwise?

The answer is yes.

Indeed, let $P_n$ ($n=1,2,\ldots$) and $P$ be any probability distributions supported on a finite interval $[a,b]$ such that $LP_n\to LP$ pointwise on an interval $[0,T]$ for some real $T>0$, where $LQ$ is the Laplace transform of a probability distribution $Q$, so that $$(LQ)(t):=\int e^{-tx}Q(dx)$$ for real $t$.

Take any subsequence $(P_{n_k})$ of the sequence $(P_n)$ that converges weakly to a probability distribution $D$. Since $e^{-tx}$ is continuous in $x$, it follows that $(LP_{n_k})(t)\to (LD)(t)$ for all real $t$. So, $$LP=LD \text{ on } [0,T]. \tag{10}\label{10}$$

Since the distribution $P$ is supported on a finite interval, it is determined by its moments $m_k(P)$ ($k=1,2,\ldots$) -- see e.g. Theorem 30.1 in Billingsley. But $$m_k(P)=(-1)^k (LP)^{(k)}(0)$$ for $k=0,1,\dots$, so that the $m_k(P)$'s are in turn determined by the values of $LP$ on $[0,T]$. So, $P$ is determined by the values of $LP$ on $[0,T]$. So, by \eqref{10}, $D=P$. So, any weakly convergent subsequence of the sequence $(P_n)$ weakly converges to $P$.

Also, since all the $P_n$'s are supported on a finite interval, the sequence $(P_n)$ is tight.

So, by the Corollary to Theorem 29.3 on p. 381 in Billingsley's book, $(P_n)$ weakly converges to $P$.

That is, the c.d.f. $C_n$ of $P_n$ converges to the c.d.f. $C$ of $P$ at every point of continuity of $C$.

Finally, if $C$ is absolutely continuous (or just continuous), then $C_n\to C$ pointwise. $\quad\Box$

$\newcommand\ep\varepsilon$The answer is yes. Indeed, we have $L(f_n)\to L(f)$ in $L^2$ as $n\to\infty$, where $L$ denotes the Laplace transform. For real $\ep>0$, let $$f_{n,\ep}:=f_n*\psi_\ep,\quad f_{\ep}:=f*\psi_\ep,$$ where $\psi_\ep$ is the p.d.f. of the mean-zero normal distribution with variance $\ep^2$ (here we extend $f_n$ and $f$ to $\Bbb R$ by $0$). Then $$L(f_{n,\ep})=L(f_n)L(\psi_\ep)\to L(f)L(\psi_\ep)=L(f_\ep)$$ in $L^2$ as $n\to\infty$. So, inverting the Laplace transform, we get $f_{n,\ep}\to f_{\ep}$ pointwise as $n\to\infty$, which implies $C_{n,\ep}\to C_{\ep}$ pointwise and hence in $L^2$ as $n\to\infty$, where $C_{n,\ep}$ and $C_{\ep}$ are the c.d.f.'s corresponding to the p.d.f.'s $f_{n,\ep}$ and $f_{\ep}$.

Also, $C_\ep\to C$ and $C_{n,\ep}\to C_n$ in $L^2$ uniformly in $n$ as $\ep\downarrow0$. Thus, $C_n\to C$ in $L^2$.

$\newcommand\ep\varepsilon$

So the question can be restated as follows: if the Laplace transforms converge pointwise, do the CDFs also converge pointwise?

The answer is yes. 

Indeed, let $P_n$ ($n=1,2,\ldots$) and $P$ be any probability distributions supported on a finite interval $[a,b]$ such that $LP_n\to LP$ pointwise on an interval $[0,T]$ for some real $T>0$, where $LQ$ is the Laplace transform of a probability distribution $Q$, so that $$(LQ)(t):=\int_{[a,b]}e^{-tx}Q(dx)$$ for real $t$.

Take any subsequence $(P_{n_k})$ of the sequence $(P_n)$ that converges weakly to a probability distribution $D$. Since $e^{-tx}$ is continuous in $x$, it follows that $(LP_{n_k})(t)\to (LD)(t)$ for all real $t$. So, $$LP=LD \text{ on } [0,T]. \tag{10}\label{10}$$

Since the distribution $P$ is supported on a finite interval, it is determined by its moments $m_k(P)$ ($k=1,2,\ldots$) -- see e.g. Theorem 30.1 in Billingsley. But $$m_k(P)=(-1)^k (LP)^{(k)}(0)$$ for $k=0,1,\dots$, so that the $m_k(P)$'s are in turn determined by the values of $LP$ on $[0,T]$. So, $P$ is determined by the values of $LP$ on $[0,T]$. So, by \eqref{10}, $D=P$. So, any weakly convergent subsequence of the sequence $(P_n)$ weakly converges to $P$.

Also, since all the $P_n$'s are supported on a finite interval, the sequence $(P_n)$ is tight.

So, by the Corollary to Theorem 29.3 on p. 381 in Billingsley's book, $(P_n)$ weakly converges to $P$.

That is, the c.d.f. $C_n$ of $P_n$ converges to the c.d.f. $C$ of $P$ at every point of continuity of $C$.

Finally, if $C$ is absolutely continuous (or just continuous), then $C_n\to C$ pointwise. $\quad\Box$