I just learned about the notion of tempered distributions $\mathcal{S}'(\mathbb{R})$.
But it is unclear that if such a distribution has a 0 derivative (of course in the distribution sense) then it equals a constant function.
Similarly, is it true that if a distribution solves Laplace equation, then it must equal to a harmonic polynomial?
Yes, zero derivative implies constant distribution. One way to see this is by mollification: if $T'=0$, then for any smooth mollifier $\phi$ we have $$(T*\phi)' = T'*\phi = 0 $$ Since $T*\phi$ is a smooth function, it's identically constant. Then the distributional limit is constant of $T*\phi_\epsilon$ as $\epsilon\to 0$ is constant too. (Here $\phi_\epsilon(t) = \epsilon^{-1}\phi(\epsilon^{-1} t)$, and one has to check that $T*\phi_\epsilon$ is independent of $\epsilon$.)
Another approach, which takes advantage of one-dimensional situation, is shown in the question here: Solving $ T' = 0 $ for distributions in $\mathbb{R}^n$. (There is no answer, but the question solves the 1-dimensional case.)
As for Laplacian: yes, you are correct. This is called Weyl's lemma (for the Laplace equation) and one of the proofs proceeds by mollification, as above. Again, $T*\phi_\epsilon$ is independent of $\epsilon$ by virtue of the mean value property, provided we use a radially symmetric mollifier.