One direction is immediate.
Suppose we assume that $\forall \varepsilon >0,\,\exists N_1\in \mathbb N$ such that $\forall n\in \mathbb N,\,n>N_1,\, |s_n-s|<\varepsilon$. So, choose $\epsilon>0$ and let $N_1$ be the number given by our assumption. Then if $n> N_1$, obviously you have $n+k > N_1$, and hence $|s_{n+k}-s|<\varepsilon$ or in other words, $|t_n-s|<\varepsilon$. Hence, under the given assumption we have $\forall \varepsilon >0,\,\exists N_2\in \mathbb N$ such that $\forall n\in \mathbb N,\,n>N_2,\, |t_n-s|<\varepsilon$.
(Note that the particular label for $N_2$ doesn't matter, we could equally well have written $\forall \varepsilon >0,\,\exists \aleph \in \mathbb N$ such that $\forall n\in \mathbb N,\,n>\aleph,\, |t_n-s|<\varepsilon$, or any other unambiguous symbol we want. The same applies to all symbols used, modulo readability.)
Now suppose $\forall \varepsilon >0,\,\exists N_2\in \mathbb N$ such that $\forall n\in \mathbb N,\,n>N_2,\, |t_n-s|<\varepsilon$. This can be rewritten as $\forall \varepsilon >0,\,\exists N_2\in \mathbb N$ such that $\forall n\in \mathbb N,\,n>N_2,\, |s_{n+k}-s|<\varepsilon$. As usual, let $\epsilon >0$. Now choose $N_1 = N_2+k$, and note that if $m>N_1$, we have $m-k >N_2$ and hence $|s_m -s| < \varepsilon$. Hence we have $\forall \varepsilon >0,\,\exists N_1\in \mathbb N$ such that $\forall n\in \mathbb N,\,m>N_1,\, |s_m-s|<\varepsilon$, or in the symbols you used initially, $\forall \varepsilon >0,\,\exists N_1\in \mathbb N$ such that $\forall n\in \mathbb N,\,n>N_1,\, |s_n-s|<\varepsilon$.