I want to find solution of the equation $T' +aT =0$ where $T \in D'(\mathbb{R})$ and $a \in \mathbb{R}$.
What I have done so far:
$$\frac{d}{dx}(e^{ax}T) = a e^{ax}T + e^{ax} \frac{dT}{dx} = e^{ax} \left(aT + \frac{dT}{dx} \right) = 0$$
This means that the distribution $e^{ax}T$ is induced by some constant $c$.
It is true that if the distributional derivative of a distribution $f$ is $0$ then $f(\psi) = c \int \psi(x) dx$ for some $c$ (see e.g. the answer to this question).
As a result, we know that $T(e^{ax} \psi) = c \int \psi(x) dx$ for each compactly supported test function $\psi$. If $\psi$ is a compactly supported test function then so is $e^{-ax}\psi$. Therefore, we see that $$T(\psi) = T(e^{ax} e^{-ax} \psi) = c \int e^{-ax} \psi(x) dx$$ so that $T$ is the regular distribution induced by $c e^{-ax}$ as desired (note that you can't hope to get rid of $c$ since the differential equation is linear and has no additional data).