**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 $$\ln\varphi_{\mu^{\ast1/n}}(t)={\rm i}tb^{(n)}-\frac{\left|\sigma^{(n)}\right|^2}2t^2+\int e^{{\rm i}tx}-1-1_{(-1,\:1)}(x){\rm i}tx\:\nu^{(n)}({\rm d}x)\tag j,$$ 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 can show that for all $\varepsilon_1,\varepsilon_2>0$, there are $k_0,n_0\in\mathbb N$ with $$\operatorname P\left[\left|b^{(n)}+\sigma^{(n)}Z+\sum_{k>k_0}\left(X_k^{(n)}-\operatorname E\left[X_k^{(n)}\right]\right)+\sum_{k=1}^{k_0}\operatorname E\left[X^{(n)}_k\right]\right|\ge\varepsilon_1\right]\le\frac{\varepsilon_2}n\tag6$$ 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 $\nu(\{a\})=\nu(\{b\})=0$.
>
> The idea is to show that $(l)$ holds for $Y^{(n)}$ replaced by $W^{(n)}$ and to use $(k)$ to verify that this already yields $(l)$.
>
> **I would highly appreciated if someone could fill these gaps.**

---

$^1$ 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 $k\ne0$).