$\newcommand{ \reals }{ \mathbb{R}} $ $\newcommand{ \distfns }{ \mathcal{D}'(\reals) } $ $\newcommand{ \testfns }{ \mathcal{D}( \reals ) } $ $\newcommand{ \ints }{ \mathbb{Z}} $ $\newcommand{ \x }{ \varphi }$
Let $\testfns$ denote $C^\infty$ functions with compact support.
Let $\distfns$ denote the space continuous linear forms on $\testfns$ such that if $\x_n \to \x$ in $\testfns$ then $\langle T, \x_n \rangle \to \langle T, \x \rangle$.
Given $\x \in \testfns$. How should one interpret the following, for $T \in \distfns$
$$ S=\sum_{m \in \ints} \tau_m T $$
where $\tau_m$ is the translation operator. Is it $ \langle S, \x \rangle = \sum_{m \in \ints} \langle \tau_m T, \x \rangle$?
Special case. Suppose that $T$ is compactly supported.
Here we have : $$\langle S, \phi \rangle = \sum_{n \in \mathbb{Z}} \langle \tau_n T , \phi \rangle = \sum_{n \in \mathbb{Z}} \langle T, \tau_{-n} \phi \rangle = \sum_{n \in \mathbb{Z}} \langle T, \phi ( \cdot - n) \rangle $$ But you have to prove this is a distribution. In this way, set $K \subset \mathbb{R}$ a compact set. Then set $\phi \in \mathcal{D}( \mathbb{R})$ compactly supported in $K$ and note that : $$\langle S, \phi \rangle = \sum_{n= -N}^N \langle T, \phi ( \cdot - n) \rangle$$ With $N$ an integer. The sum is truncated because when $n$ is large inough, the suport of $\phi ( \cdot -n)$ has no intersection with the support of $T$.
But $\|\tau_{-n} \phi\|_{\mathcal{C}^p} = \|\phi\|_{ \mathcal{C}^p}$ for every $p$ so that since $T$ is a distribution there existe $C_K >0$ and $p_K \geqslant 0$ such that : $| \langle T, \phi \rangle | \leqslant C_K \|\phi\|_{\mathcal{C}^{p_K}}$ where $C_K,p_K$ only depend on $K$, not $\phi$. Thus we obtain : $$\langle S, \phi \rangle \leqslant (2N+1) C_K \|\phi\|_{\mathcal{C}^{p_K}}$$
This shows that $S$ is a distribution.
More generally I think that letting $\displaystyle S_N:= \sum_{n=-N}^N \tau_n T$ you have to figure out if $(S_N)_{N \geqslant 0}$ converges or not.
For example if you set $\displaystyle \langle T, \phi \rangle = \int_{\mathbb{R}} \phi(x) \mathrm{d}x$ then $S$ is not defined :
Let $\phi \in \mathcal{D}(\mathbb{R})$ such that $\displaystyle \int_{\mathbb{R}} f(x) \mathrm{d}x = 1$ then : [\tau_n T=T] for all $n \in \mathbb{Z}$ thus $\langle S_N, \phi \rangle = 2N+1$ and does not converge.