The transformation $\theta$ on $\Omega^{\Bbb N}$ is ergodic. Indeed, it's enough to show that for each cylinder $A$ and $B$, we have
$$\frac 1n\sum_{k=0}^{n-1}\mu(\theta^{-k}A\cap B)\to \mu(A)\mu(B),$$
where $\mu$ is the measure on the product $\sigma$-algebra.
If $A=\prod_{j=0}^NA_j\times \Omega\times\dots$ and $B=\prod_{j=0}^NB_j\times \Omega\times\dots$, we have for $k>N$
\begin{align}
\theta^{-k}A\cap B&=\{(x_j)_{j\geq 0}, (x_{j+k})_{j\geq 0}\in A, (x_j)_{j\geq 0}\in B\}\\
&=\{(x_j)_{j\geq 0},x_{j+k}\in A_j, 0\leq j\leq N, x_j\in B_j,0\leq j\leq N\}\\
&=B_0\times \dots\times B_N\times \Omega\times\dots\times \Omega\times A_0\times\dots\times A_n\times \Omega\times\dots,
\end{align}
and we use the definition of product measure $\mu$ on cylinders (the $N$ first terms doesn't matter).
Since $\theta$ is ergodic, $\mathcal J_{\theta}$ consists only of events of measure $0$ or $1$. The conditional expectation with respect such a $\sigma$-algebra is necessarily constant.
Let me expand here the point made in the the comments. First, I misread slightly your question. I had in mind two-parametric ergodic means given by
$$
A_{n,m}(f) = \sum_{i = 0}^{n - 1} \sum_{j = 0}^{m - 1} f(T^i x, T^j y).
$$
Not the means $A_{n}(f) = A_{n,n}(f)$. For the first you need to be in $L \log L(\Omega)$. For the second I do not have a definitive answer. It may be enough to be in $L^1$, see the edit bellow.
First, the following result is known.
Theorem: For every $f \in L \log L(\Omega,\mu)$ , where $(\Omega, \mu) = (\Omega_1 \times \Omega_2, \mu_1 \otimes \mu_2)$ and ergodic transformation $T_i:X_i \to X_i$, the two parametric ergodic averages
$$
A_{n,m}(f) = \sum_{i = 0}^{n - 1} \sum_{j = 0}^{m - 1} f(T^i x, T^j y)
$$
converge almost everywhere $A_{n,m}(f) \to f$.
The proof uses the following steps
- For each $i$ use the maximal weak-type $(1,1)$ maximal ergodic inequality.
- Use real interpolation to get that:
$$
\Big\| \sup_{n \geq 0} A^i_{n}(|f|) \Big\|_p \lesssim \max \Big\{ 1, \frac1{p - 1} \Big\} \, \| f \|_p
$$
- use Yano's extrapolation [Y] to obtain that the above maximal operator above maps $L \log L(\Omega)$ into $L^1(\Omega)$ for every $i$.
Then, copying the argument in [JMZ], you have that the map $f \mapsto f^\ast$ given by
$$
f^\ast(x,y) = \limsup_{n, m \to \infty} A_{n,m}(|f|)
$$
maps $L \log L(\Omega)$ into $L_1(\Omega)$. Just by composition, we have that
$$
\limsup_{n, m \to \infty} A_{n,m}(|f|)
\leq
\limsup_{n \to \infty} A^1_{n} \bigg( \underbrace{\sup_{m \geq 0} A_m^2|f|}_{g} \bigg)
$$
But the function $g$ is in $L^1(\Omega)$ as the function $f$ is in $L \log L$. Now, using that for every function in $L^1(\Omega)$ we have almost everywhere convergence (and therefore the limsup is exchangeable by the lim) we can conclude. The fact that $f^\ast$ is in $L^1$ gives almost everywhere convergence by the same argument that is used with maximal functions.
I will go further and conjecture that the following is true (probably known to the experts):
Open (to my knowledge) For every $\varphi$ with $\varphi \in o(x \log x)$ we have that there is an element $f \in L_\varphi(\Omega)$, such that $A_{n,m}(f) \not\to f$.
It is known, see [JMZ, Theorem 8]. That this holds in the case of differentiability of integrals. I will try to adapt the argument to the ergodic case. It is also likely to be true, since something similar holds for martingales [G].
[G] Gundy, R. F., On the class L log L, martingales, and singular integrals, Stud. Math. 33, 109-118 (1969). ZBL0181.44202.
[JMZ] Jessen, B.; Marcinkiewicz, J.; Zygmund, A., Note on the differentiability of multiple integrals., Fundamenta Math. 25, 217-234 (1935). ZBL61.0255.01.
[Y] Yano, Shigeki, Notes on Fourier analysis. XXIX. An extrapolation theorem, J. Math. Soc. Japan 3, 296-305 (1951). ZBL0045.17901.
P.D.: For your purposes, perhaps it will be enough if you take the usual ergodic averages with respect to
$$
S = p (\mathrm{id} \otimes T) + q (T \otimes \mathrm{id})
$$
for $p + q = 1$. Its ergodic averages will be weighted summations concentrated as Gaussian around the diagonal $i = j$.
Edit I found a (almost) complete solution in $L^1$ in the following paper:
You need to interpret $(i,j) \mapsto T^i \otimes T^j$ as an action of
$\mathbb{Z}^2$ and use that $[0,N]$ is a well-tempered Foelner
sequence (in the sense of that paper). That would give you almost
everywhere convergence for the means $A_{n,n}(f)$, for every $f \in
L^1(\Omega)$.
Best Answer
Your choice for $(a,b)$ does not give rationally independent elements, because $0\cdot a+1\cdot b=0$ (with $n=0$, $m=1$) hence the transformation is not ergodic.