Apart from some quibbling about making sure $K$ is sufficiently integrable or something... this is true. E.g., for precision, take $K$ to be a tempered distribution in two variables. Using the hypothesis that $2\pi ix\cdot T=T\cdot {d\over dx}$ as operators on Schwartz functions, (thinking of "$x$" as multiplication-by-$x$), integration by parts in the integral for $T$ gives ${\partial\over \partial y}K(x,y)=2\pi i x\cdot K(x,y)$ as tempered distribution in two variables. This has obvious classical solutions $C\cdot e^{2\pi ixy}$, as expected. To show that there are no others, among tempered distributions, one way is to divide $K(x,y)$ by $e^{2\pi ixy}$, so the equation becomes ${\partial \over \partial y}K(x,y)=0$. By symmetry, ${\partial\over \partial x}K(x,y)=0$. Integrating, $K(x,y)$ is a translation-invariant tempered distribution in two variables. It is a separate exercise to see that all such are constants.

Edit: in response to @GiuseppeNegro's comment (apart from correcting the sign), the secondary exercise of proving that vanishing first partials implies that a tempered distribution is (integrate-against-) a constant has different solutions depending on one's context, I think. Even in just a single variable, while we can instantly invoke the mean value theorem to prove that a function with pointwise values is constant when it is differentiable and has vanishing derivative, that literal argument does not immediately apply to distributions. In a single variable, integration by parts proves that $u'=0$ for distribution $u$ implies that $u(f')=0\,$ for all test functions $f$, *and* we can characterize such $f$, namely, that their integrals over the whole line are $0$, from which a small further argument proves that $u$ is a constant. This sort of argument seems to become a little uglier in more than one variable... and, in any case, I tend to favor a slightly different argument that is a special case of proving uniqueness of various group-invariant functionals. E.g., on a real Lie group $G$, there is a unique right $G$-invariant distribution (=functional on test functions), and it is integration-against right Haar measure. The argument is essentially just interchange of the functional and integration against an approximate identity, justified in the context of Gelfand-Pettis (weak) integrals. Probably there are more elementary arguments, but this sort of approach seems *clearer* and more persuasive in the long run.

Edit-Edit: on a Lie group $G$, to prove that all distributions annihilated by the left $G$-invariant differential operators attached to the Lie algebra $\mathfrak g$ (acting on the right) are (integrate-against-) constants: Let $f_n$ be a Dirac sequence of test functions. A test function $f$ acts on distributions by $f\cdot u=\int_G f(g)\,R_gu\;dg$, where $R$ is right translation, and the integral is distribution-valued (e.g., Gelfand-Pettis). A basic property of vector-valued integrals is that $f_n\cdot u\to u$ in the topology on distributions. At the same time, the distribution $f_n\cdot u$ is (integration-against) a smooth function. It is annihilated by all invariant first-order operators, so by the Mean Value Theorem it is (integration against) a constant. The distributional limit of constants is a constant.