I find another proof using functional calculus and spectral theory. Denote x=ST, y=TS. Let f(t)=1+t, then x=f(y). Therefore, Sp(x)=Sp(f(y))=f(Sp(y))=1+Sp(y)——-(1) On the other hand, by [Completed Normed Algebras, F.F. Bonsall J.Duncan] P.20, Proposition 3, Sp(x)\{0}=Sp(y)\{0} ——– (2) (1) and (2) are contradicted to each other since Sp(x) and Sp(y) are both non-empty compact sets.

(A non-empty compact set cannot be translation invariant)