If signed measures $\mu_n$ are such that $\mu_n\to\mu$ and $\|\mu_n\|\to c\in(0,\infty)$, does $\exp^*(\mu_n)/\|\exp^*(\mu_n)\|$ necessarily converge?

$$\newcommand{\R}{\mathbb R}$$Let $$M$$ denote the set of all finite signed measures on a separable Banach space $$B$$. For any $$\mu\in M$$, let $$\begin{equation*} \exp^*(\mu):=\sum_{k=0}^\infty\frac{\mu^{*k}}{k!}. \end{equation*}$$ The following question arose during a discussion on this page:

Suppose that $$(\mu_n)$$ is sequence in $$M$$ such that $$\mu_n\to\mu$$ (weakly) for some $$\mu\in M$$ and $$\|\mu_n\|\to c$$ for some $$c\in(0,\infty)$$, where $$\|\cdot\|$$ is the total variation norm. Does it then follow that the sequence $$(\nu_n)$$ with $$\begin{equation*} \nu_n:=\frac{\exp^*(\mu_n)}{\|\exp^*(\mu_n)\|} \end{equation*}$$ converges?

This question will be answered, negatively, below.

$$\newcommand{\R}{\mathbb R}\newcommand{\la}{\lambda}\newcommand{\de}{\delta}\newcommand{\ka}{\rho}$$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 $$\begin{equation*} b:=\|\exp^*(\ka_n)\|\le\|\ka_n^{*0}+\ka_n^{*1}\| +\sum_{k=2}^\infty\frac{\|\ka_n\|^k}{k!}=2+(e^3-1-3)=e^3-2 and hence, in view of (8), $$\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.