How strong is an oracle that avoid don't-halt
EDIT: I made a deeply silly mistake in my previous answer. Below what I've done is answered the question, then explained how two variations of the question - which I originally conflated with the actual question - yield a very different answer.



An oracle of the type you describe computes the halting problem (and conversely every oracle computing the halting problem computes an oracle of your type).


Suppose $x$ computes the halting problem. Without loss of generality, we may take $x$ to literally be the halting problem. Then we can compute an $f$ as desired by simply taking an input $e$, asking $x$ whether $T_e(e)$ halts, and (i) if the answer is no, halt and output $0$ and (ii) if the answer is yes, run $T_e$ until it halts and output the result.
Conversely, suppose $f$ has the desired property. There is a total computable function $h$ such that for all $e$ if $T_e(e)$ halts, then $T_{h(e)}(h(e))$ halts and gives the runtime of $T_e(e)$. Now to tell whether $T_e(e)$ halts we simply run $T_e(e)$ for $f(h(e))$-many steps.




Now there are a couple natural variations to the above notion:


What if we demand inequality instead of equality?
What if we restrict attention to machines with binary output?


Both of these variations prevent the argument above from working, the former by preventing us from representing runtimes as outputs and the latter by preventing us from coding literally anything into the output.

It turns out that these notions yield a much weaker complexity class: namely the PA degrees. The original definition was that something is of PA degree if it computes a complete consistent extension of Peano arithmetic (which by Godel must be non-computable), but there are many equivalent definitions. Generally the simplest to work with is:


  A degree ${\bf a}$ is PA iff whenever $X$ is an infinite computable binary tree, ${\bf a}$ computes an infinite path through $X$.


PA degrees can be quite weak, computability-theoretically. While no PA degree is computable, they don't have to compute the halting problem: there are PA degrees incomparable with the halting problem, and there are PA degrees strictly weaker than the halting problem (where strength here is measured by Turing reducibility).



Re: relativization, the answer is no: the proof of the incomputability of all PA degrees relativizes, so no oracle is PA over itself.