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
$$
T^n = \frac{1}{2^n} (A + B)^n = \frac{1}{2^{n}} \sum_{k=0}^{n} \binom{n}{k} A^{k} B^{n-k}
= \frac{1}{2^{n}} B^{n} \sum_{k=0}^{n} \binom{n}{k} C^{k},
$$
where $C = AB^{-1}$ so that $Cf(x) = f(x + a - b)$. It think that one should be able to show that this converges 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 ...