Here is my reasoning. I use multiplicative notation.
Let $f \in L^2$ then it has a Fourier series $$f(x,y)=\sum a_{n,m}x^ny^m$$ because $x^ny^m$ are characters and so they form an orthonormal basis. Now $$f(T(x,y))=\sum a_{n,m}x^{n+m}y^m.$$ If $f$ is $T$ invariant then it must be that $a_{n,m}=a_{n+m,m}=a_{n+2m,m}=...$ Thus unless $a_{n,m}$ are zero this the coefficents are not square summable. The series $$\sum a_{n,m}x^{n+m}y^m$$ is a Fourier series as $x^{n+m}y^m$ are disinct charcters for all $n,m$.
Where is my mistake? My books says this map is not ergodic.
Edit: Is $m=0$ perhaps the problem here? since $a_{n,0}$ might still be non-zero. If so, I demand a refund of my time i spent staring at this
Assuming you mean with respect to Lebesgue measure: $f(x,y) = x$ is a non-constant $T$-invariant function.
In the Fourier series for this $f$ you have all $a_{n,m}=0$ except for $a_{1,0}=1$. So you were correct when you guessed that the coefficients with m=0 create the problem with your original proof