Let $(X_n)_n$ be an i.i.d sequence of random variables defined on the same probability space and denote $S_n=\frac{1}{n}\sum_{k=1}^n X_k.$
Assume that $\mathbb{E}(X_1)\ne0$ and given $\alpha>0$ that $$\sqrt{n}(X_n-\alpha\mathbb{1}_{\{S_n\ne 0\}})\overset{d}{\to} \frac{X_1}{\mathbb{E}(X_1)}.$$
I would like to deduce that $\sqrt{n}(X_n-\alpha)$ converge in distribution. My idea is to use the strong long of large numbers that $\mathbb{1}_{\{S_n\ne 0\}}\overset{\text{a.e.}}\to 1$ but not sure how can I conclude.
Write $$\sqrt{n}(X_n-\alpha)=\sqrt{n}(X_n-\alpha \mathbb{1}_{\{S_n\ne 0\}})+\sqrt{n}\alpha(\mathbb{1}_{\{S_n\ne 0\}}-1).$$ Since $\mathbb{1}_{\{S_n\ne 0\}}$ can take values $0\vee 1$ only, convergence $\mathbb{1}_{\{S_n\ne 0\}}\overset{\text{a.e.}}\to 1$ means that for every elementary event $\omega\in A$, where $\mathbb P(A)=1$, there exists $N=N(\omega)$ s.t. for all $n\geq N$ $$ \mathbb{1}_{\{S_n\ne 0\}}(\omega):=\mathbb{1}_{\{S_n(\omega)\ne 0\}} = 1. $$ So, for every $\omega\in A$ and for each $n\geq N$ the additional term vanishes: $$\sqrt{n}\alpha(\mathbb{1}_{\{S_n\ne 0\}}-1)(\omega)=0.$$ We proved that $$ \sqrt{n}\alpha(\mathbb{1}_{\{S_n\ne 0\}}-1)\overset{\text{a.e.}}\to 0. $$ Next use Slutsky theorem as proposed by anonymus.