I'm trying to prove the following equation:
For distributions $S$ and $T$ on the space of test functions:
$e^{\lambda x} (T \circledast S) = e^{\lambda x} T \circledast e^{\lambda x} S$
What I have is:
$<e^{\lambda x}(T \circledast S), \phi (x)> \\ = <T_{x}, <S_{y}, e^{\lambda (x+y)} \phi (x+y)>> \\ =<e^{\lambda x} T_{x}, <e^{\lambda y}S_{y}, \phi (x+y)>> \\ =<e^{\lambda x} T \circledast e^{\lambda x} S, \phi (x)> \\$
But I'm not sure wether this is a legit proof...
You need more hypotheses! In general the convolution of two distributions does not even exist.
The second line in the display in your post is, or at least may be, meaningless. The problem is that the function $$x\mapsto<S_{y}, e^{\lambda (x+y)} \phi (x+y)>$$need not be a test function (for example there's no reason to think it has compact support).