$\newcommand{\R}{\mathbb R}\newcommand{\la}{\lambda}\newcommand{\de}{\delta}\newcommand{\ka}{\rho}$Let $B=\R$.

For each natural $n$, let
\begin{equation*}
\la_n:=2\de_{\pi/n}-\de_{1/n},\tag{1}
\end{equation*}
where $\de_a$ is the Dirac probability measure at point $a$. Then
\begin{equation*}
\la_n\to\mu:=\de_0 \tag{2}
\end{equation*}
and hence, by dominated convergence or using characteristic functions,

\begin{equation*}
\exp^*(\la_n)\to\exp^*(\mu)=\exp^*(\de_0)=e\de_0. \tag{3}
\end{equation*}
Also, for each $k=0,1,\dots$ we have
$$\la_n^{*k}=\sum_{j=0}^k\binom kj2^j(-1)^{k-j}\de_{j\pi/n-(k-j)/n};$$
therefore and because the values of $j\pi/n-(k-j)/n=j\big(\pi-(-1)\big)/n-k/n$ are distinct for distinct values of $j\in\{0,\dots,k\}$, we have

\begin{equation*}
\|\la_n^{*k}\|=\sum_{j=0}^k\binom kj2^j=3^k. \tag{4}
\end{equation*}
Since $\pi$ is irrational, it is easy to see that for each $n$ the signed measures $\la_n^{*0},\la_n^{*1},\la_n^{*2},\dots$ are mutually singular, whence
\begin{equation*}
\|\exp^*(\la_n)\|=\sum_{k=0}^\infty\frac{\|\la_n^{*k}\|}{k!}=\sum_{k=0}^\infty\frac{3^k}{k!}=e^3. \tag{5}
\end{equation*}
Thus, by (3),
\begin{equation*}
\frac{\exp^*(\la_n)}{\|\exp^*(\la_n)\|}\to\frac{e\de_0}{e^3}. \tag{6}
\end{equation*}

Somewhat similarly to (1), for each natural $n$, let
$$\ka_n:=2\de_{1/n}-\de_0.$$
Then, quite similarly to (2)--(4), we have
\begin{equation*}
\ka_n\to\mu=\de_0,\tag{7}
\end{equation*}
\begin{equation*}
\exp^*(\ka_n)\to\exp^*(\mu)=e\de_0,\tag{8}
\end{equation*}
\begin{equation*}
\|\ka_n^{*k}\|=3^k.
\end{equation*}

However, (5) and (6) do not hold with $\ka_n$ in place of $\la_n$. Indeed,
\begin{equation*}
\|\ka_n^{*0}+\ka_n^{*1}\|=\|\de_0+(2\de_{1/n}-\de_0)\|=2,
\end{equation*}
so that
\begin{equation*}
b:=\|\exp^*(\ka_n)\|\le\|\ka_n^{*0}+\ka_n^{*1}\|
+\sum_{k=2}^\infty\frac{\|\ka_n\|^k}{k!}=2+(e^3-1-3)=e^3-2<e^3
\end{equation*}
and hence, in view of (8),
\begin{equation*}
\frac{\exp^*(\ka_n)}{\|\exp^*(\ka_n)\|}\to\frac{e\de_0}b,\quad\text{and}\quad b\ne e^3; \tag{9}
\end{equation*}
(since $\exp^*(\ka_n)(A)=\exp^*(\ka_1)(nA)$ for all $A\subseteq\mathbb R$, we see that $b=\|\exp^*(\ka_n)\|$ does not depend on $n$).

Let now $\mu_n:=\la_n$ if $n$ is odd and $\mu_n:=\ka_n$ if $n$ is even. Then, in view of (2), (7), (3), (8),
\begin{equation*}
\mu_n\to\mu
\end{equation*}
and
\begin{equation*}
\exp^*(\mu_n)\to\exp^*(\mu).
\end{equation*}
Also,
\begin{equation}
\|\mu_n\|=3\to3.
\end{equation}
However, in view of (6) and (9),
\begin{equation*}
\frac{\exp^*(\mu_n)}{\|\exp^*(\mu_n)\|}
\end{equation*}
does not converge.