Why are translation invariant operators on $L^2$ multiplier operators

Here's a sketch proof, which I think gets to the core reason why this is true. Remember that the fourier transform takes translations to multiplication by characters. So if $T$ commutes with translation, $\hat T = \mathcal{F} T \mathcal{F}^{-1}$ will commute with multiplication by characters. So $\hat T$ commutes with all operators which are in the closure of the linear span of the operators given by multiplication by characters. That is, $\hat T$ commutes with multiplication by any continuous function. It's not so hard to then show that $\hat T$ must itself by multiplication by some $m\in L^\infty$; under your definition, this means precisely that $T = T_m$.

This is Theorem 3.16 in the first chapter of Introduction to Fourier Analysis on Euclidean Spaces, by E.M. Stein and G. Weiss.