$\newcommand\ep\varepsilon$
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. $\quad\Box$