I would like to show that $$\lim_{h\to 0}\frac{\tau_{he_j}T-T}{h}=\partial_{x_j}T$$ for every distribution $T\in\mathcal{D}'(\mathbb{R})$, where $\tau_{he_j}\varphi(x)=\varphi(x-he_j)$ and $e_j$ is an element of the canonical basis of $\mathbb{R}^n$.
Taking a function $\varphi\in C_0^\infty(\mathbb{R})$, we have $$\lim_{h\to 0}\left\langle\frac{\tau_{he_j}T-T}{h},\varphi\right\rangle=\lim_{h\to 0}\left\langle T,\frac{\tau_{-he_j}\varphi-\varphi}{h}\right\rangle,$$ but the continuity is only guaranteed over $C_K^\infty(\mathbb{R})$ for $K$ compact, so I don't know how to exchange the limit and the distribution.
For distributions acting on $\mathcal{D}(\Omega)=\mathcal{C}^\infty_c(\Omega)$ (smooth functions with compact support in $\Omega$, one may appeal to the mean value theorem.
Since $$\lim_{h\to 0}\left\langle\frac{\tau_{he_j}T-T}{h},\varphi\right\rangle=\lim_{h\to 0}\left\langle T,\frac{\tau_{-he_j}\varphi-\varphi}{h}\right\rangle,$$ It is enough to show that $h^{-1}(\tau_{he_j}\phi-\phi)$ converges to $\partial_{x_j}\phi$ in $\mathcal{C}^\infty_c(\Omega)$ for any $\phi\in\mathcal{C}^\infty_c(\Omega)$.
The mean value theorem implies that every $\psi\in\mathcal{D}(\Omega)$ is Lipschitz. Another application of the mean value theorem show that there is a constant $C=C(\phi,\alpha)$ such that \begin{align*} \Big|\frac{\partial^\alpha\phi(x)-\partial^\alpha\phi(x-he_j)}{h} - \partial_{x_j}\partial^\alpha\phi(x)\Big| &=|\partial_{x_j}\partial^\alpha\phi(x-t\theta h)-\partial_{x_j}\partial^\alpha\phi(x)|\\ &\leq C|h|\xrightarrow{h\rightarrow0}0 \end{align*} where $\theta=\theta(h,x)\in (0,1)$.
For distributions acting on the Schwartz space, the proof is similar but a little more involved.