Partial answer
Let
- $(b,\sigma)\in\mathbb R\times[0,\infty)$;
- $\nu$ be a $\sigma$-finite measure on $\mathbb R$ with $$\int1\wedge x^2\:\nu({\rm d}x)<\infty\tag a$$ and $\nu(\{0\})=0$;
- $\mu$ be a probability measure on $\mathbb R$ with $$\ln\varphi_\mu(t)={\rm i}tb-\frac{\sigma^2}2t^2+\int e^{{\rm i}tx}-1-1_{(-1,\:1)}(x){\rm i}tx\:\nu({\rm d}x)\tag b$$ for all $t\in\mathbb R$.
We can construct a real-valued random variable $Y$ on a probability space $(\Omega,\mathcal A,\operatorname P)$ with $Y\sim\mu$ in the following way:
- Let $$I_k:=\left(-\frac1k,-\frac1{k+1}\right]\cup\left[\frac1k,\frac1{k+1}\right)$$ and $$\nu_k(B):=\nu(B\cap I_k)\;\;\;\text{for }B\in\mathcal B(\mathbb R)$$ for $k\in\mathbb N$
- Note that $$\nu(I_0)+\int_{(-1,\:1)}\nu({\rm d}x)=\int1\wedge x^2\:\nu({\rm d}x)<\infty\tag c.$$
- Let $(X_k)_{k\in\mathbb N_0}$ be a real-valued independent process on $(\Omega,\mathcal A,\operatorname P)$ with$^1$ $$X_k\sim\operatorname{CPoi}_{\nu_k}\;\;\;\text{for all }k\in\mathbb N_0\tag d.$$
- Note that $$\operatorname E[X_k]=\int_{I_k}\nu({\rm d}x)x\tag e$$ and $$\operatorname{Var}[X_k]=\int_{I_k}\nu({\rm d}x)x^2\tag f$$ for all $k\in\mathbb N_0$.
- It's easy to see that $$M_k:=\sum_{i=1}^k\left(X_i-\operatorname E\left[X_i\right]\right)\;\;\;\text{for }k\in\mathbb N$$ is a martingale with $$\operatorname E\left[M_k^2\right]=\sum_{i=1}^k\operatorname{Var}[X_i]\tag g\;\;\;\text{for all }n\in\mathbb N$$
- Let $\mathcal F^X_\infty:=\sigma(X_k:k\in\mathbb N)$.
- Then, $$\sup_{k\in\mathbb N}\operatorname E\left[M_k^2\right]=\int_{(-1,\:1)}\nu({\rm d}x)x^2<\infty\tag h$$ and hence $$M_k\xrightarrow{k\to\infty}M_\infty\;\;\;\text{almost surely}\tag i$$ for some real-valued $\mathcal F^X_\infty$-measurable square-integrable random variable $M_\infty$ on $(\Omega,\mathcal A,\operatorname P)$ by the martingale convergence theorem.
- Now let $Z$ be a real-valued standard normally distributed random variable on $(\Omega,\mathcal A,\operatorname P)$ independent of $(X_n)_{n\in\mathbb N_0}$
- It's easy to show that $$Y:=b+\sigma Z+X_0+M_\infty\sim \mu.$$
Now we know that there characteristic exponent of $\mu^{\ast1/n}$ is similarly given by $$\ln\varphi_{\mu^{\ast1/n}}(t)={\rm i}tb^{(n)}-\frac{\left|\sigma^{(n)}\right|^2}2t^2+\int e^{{\rm i}tx}-1-1_{(-1,\:1)}(x){\rm i}tx\:\nu^{(n)}({\rm d}x)\tag j,$$ where $b^{(n)}:=b/n$, $\sigma^{(n)}:=\sigma/\sqrt n$ and $\nu^{(n)}:=\nu/n$, for all $n\in\mathbb N$. Define $\left(\nu^{(n)}_k,X^{(n)}_k,M^{(n)}_k,M^{(n)}_\infty,Y^{(n)}\right)$ in the same way as before, but with $(b,\sigma,\nu)$ replaced by $\left(b^{(n)},\sigma^{(n)},\nu^{(n)}\right)$.
We then should be able to show that for all $\varepsilon_1,\varepsilon_2>0$, there are $k_0,n_0\in\mathbb N$ with $$p_{k_0}^{(n)}:=\operatorname P\left[\left|b^{(n)}+\sigma^{(n)}Z-\sum_{k=1}^{k_0}\operatorname E\left[X^{(n)}_k\right]\right|+\sum_{k>k_0}\left(X_k^{(n)}-\operatorname E\left[X_k^{(n)}\right]\right)>\varepsilon_1\right]\le\frac{\varepsilon_2}n\tag k$$ for all $n\ge n_0$.
Now let $$W^{(n)}:=\sum_{k=0}^{k_0}X_k^{(n)}\;\;\;\text{for }n\in\mathbb N.$$
In order to conclude that $n\mu^{\ast1/n}\to\nu$ vaguely, it is sufficient to show that $$n\operatorname P\left[Y^{(n)}\in(a,b]\right]\xrightarrow{n\to\infty}\nu((a,b])\tag l$$ for all $a,b\in\mathbb R$ with $a<b$ and $\nu(\{a\})=\nu(\{b\})=0$.
Using the simple estimate $\operatorname P\left[U\in(a+\varepsilon_1,b-\varepsilon_1]\right]-\operatorname P[|V|>\varepsilon_1]\le\operatorname P[U+V\in(a,b]]\le\operatorname P[U\in(a-\varepsilon_1,b+\varepsilon_1]]+\operatorname P[|V|>\varepsilon_1]$, for every random variables $U,V$, we obtain \begin{equation}\begin{split}&\operatorname P\left[W^{(n)}\in(a+\varepsilon_1,b-\varepsilon_1]\right]-p_{k_0}^{(n)}\\&\;\;\;\;\le\operatorname P[Y^{(n)}\in(a,b]]\\&\;\;\;\;\le\operatorname P[W^{(n)}\in(a-\varepsilon_1,b+\varepsilon_1]]+p_{k_0}^{(n)}\end{split}\tag m\end{equation} for all $n\ge n_0$. On the other hand, letting $J_{k_0}:=\bigcup_{k=0}^{k_0}I_k$, $$n\operatorname P\left[W^{(n)}\in B\right]=\nu(B\cap J_{k_0})\xrightarrow{k_0\to\infty}\nu(B)\;\;\;\text{for all }B\in\mathcal B(\mathbb R)\tag n.$$ So, unless I'm missing something, we should be able to conclude as long as we can justify that we can take $k_0$ as large as desired, while maintaining $(k)$.
I would highly appreciated if someone could fill this gap.
$^1$ One issue, which I'm unable to resolve at the moment, is that $\operatorname{CPoi}_{\eta}$ is only well-defined, when $\eta$ is a finite measure on $\mathbb R$, but (unless I'm missing something) the assumptions don't imply that $\nu_k$ is finite for all $k\in\mathbb N_0$. Since $\nu$ is arising from the Lévy-Khinchin formula, we may be able (I don't know whether this is true) to assume that $\nu$ is locally finite. Under this assumption, $\nu_k$ would be finite for all $k\ne0$ (since the support of $\nu_k$ is contained in $(-1,1)$ for $k\ne0$).