Ok, I rethought my old comment. I believe it is better with $Af (x) = f(x+a)$ and $B f(x) = f(x+b)$ to think about $$ T^n = \frac{1}{2^n} (A + B)^n = \frac{1}{2^{n}} \sum_{k=0}^{n} \binom{n}{k} A^{k} B^{n-k} = \frac{1}{2^{n}} B^{n} \sum_{k=0}^{n} \binom{n}{k} C^{k}, $$ We see where $C = AB^{-1}$ so that there is an averaging as required for the convergence of ergodic averages$Cf(x) = f(x + a - b)$.
One can now It think that one should be able to show that $T^n$ this converges at least in $L^2$.relatively easily ... (one somehow needs to deal with the weights).
Old Post
Let me rephrase the answer of Fabrizio Polo first:
Consider all words $w$ in A, B of length $n$. Call this set $\mathcal{W}_n$. Now define $Af (x) = f(x+a)$ and $B f(x) = f(x+b)$. Then $T^n$ from the original post is equal to $$ \frac{1}{|\mathcal{W}_n|} \sum w, $$ where the sum is taken over all elements of $\mathcal{W}_n$. I am somehow unable to make that display properly. Here $w$ stands for the appropriate product of operators. E.g. for $n - 2$, we have $\mathcal{W}_n = \{AA, AB, BA, BB\}$ so that the expression above becomes $$ \frac{1}{4} (AA + AB + BA + BB), $$ which is the $T^2$ from the original post.
Now if $a - b$ is irrational, I believe that $(\mathbb Z_+) \ast (\mathbb Z_+)$ action defined above is ergodic, so one should have almost sure convergence. However, I am not sure if this holds, since the group $(\mathbb Z_+)\ast(\mathbb Z_+)$ is not ameanable. So you will probably have to look into ergodic theorems for non ameanable actions to answer this question.
Another hope could be to somehow resum the expression for $T^n$ and be able to use more classical ergodic theorems ...
