[Edit: I misunderstood your definition, and read as describing an injective function. The "solution" below explains why this cannot work.]
What you are suggesting is interesting. The question is whether we can assign to each limit ordinal a cofinal sequence in such a way that the $f$ you describe ends up being injective. I do not know off hand whether this is possible, but it is a nice problem (it is not clear yet whether $f$ is well-defined, i.e., whether one can arrange so that all the relevant series actually converge). Sorry for the original confusion.
The argument you are giving cannot work. This is because in your construction you are not only trying to define an injection, but in fact a strictly increasing function.
But if $f:\Omega\to{\mathbb R}$ is increasing, then for each $\alpha$ there will be rational in the interval between $f(\alpha)$ and $f(\alpha+1)$, and different values of $\alpha$ will correspond to different rationals. Of course, this is impossible as $\Omega$ is uncountable but ${\mathbb Q}$ is countable.
In fact, you will not be able to define an injection $f$ by any explicit procedure. This is because it is consistent with the axioms of set theory without choice that no such injections exist, and it is consistent with set theory plus choice that no such injection is definable.
One way to show that there are such injections is to use Zorn's lemma on the collection of injections $f$ whose domain is an ordinal. This is a partial order: $f\le g$ if $g$ extends $f$ as a function, i.e., iff $g$ has a larger domain than $f$, and the restriction of $g$ to the domain of $f$ is just $f$. Clearly a maximal element in this poset must have uncountable domain, since the reals are uncountable.
This is very easy to understand when using the order-theoretic definition of $\alpha+\beta$ as the linear order which is the initial segment $\alpha$, followed by the tail segment $\beta$.
Remember now that there is at most a single embedding of one ordinal into another whose image is an initial segment.
We note that $\gamma$ is a joint initial segment of both $\gamma+\alpha$ and $\gamma+\beta$, so the identity function is the only embedding. Now we may proceed by considering the embedding of $\alpha$ into $\beta$, which itself is also the identity, and using it to extend the embedding of $\gamma\to\gamma$.
This shows, easily, that $\gamma+\alpha\leq\gamma+\beta$. But now we remember that the embedding $\alpha\to\beta$ was not surjective, so the embedding we got is not surjective. Since that is the only embedding onto an initial segment, it must be that $\gamma+\alpha<\gamma+\beta$.
Of course, this doesn't help you if you're trying to prove the inequality from the recursive definition. But it gives a good picture as to what's going on.
Now fix $\gamma$, and prove by induction on $\beta$, that for all $\alpha<\beta$, $\gamma+\alpha<\gamma+\beta$; and conclude this is true for all $\alpha,\beta,$ and $\gamma$.
Best Answer
Edit: There's an amusing generalization of all this at the bottom.
An ordinal determines an ordering. The ordering corresponding to the sum of two ordinals is "first an ordering like this one, followed by an ordering like that one".
So for example the ordinal $3$ corresponds to three things in a row: $xxx$ .
And $\omega$ corresponds to an infinite sequence: $xxxxxx\dots$ .
If you take three things in a row and follow them by an infinite sequence you get an infinite sequence: $$xxx,xxx\dots = xxx\dots$$So $3+\omega=\omega$.
If $n<\omega$ then $n+\alpha=\alpha$ for any infinite ordinal $\alpha$, nothing to do with whether it's a limit ordinal. Because $\alpha$ starts with an infinite sequence, then perhaps has more added on - that sequence eats the $n$, as in the example $3+\omega=\omega$ above.
Amusing Generalization. For $n<\omega$ define $\alpha n=\alpha+\dots+\alpha$. Define $\alpha\omega=\bigcup_{n<\omega}\alpha n.$
Theorem Suppose $\alpha$ and $\beta$ are ordinals. We have $\alpha+\beta=\beta$ if and only if $\alpha+\beta$ is obviously equal to $\beta$.
Or a little more explicitly,
Theorem Given two ordinals $\alpha$ and $\beta$, we have $\alpha+\beta=\beta$ if and only if $\beta\ge\alpha\omega$.
Proof: If $\alpha+\beta=\beta$ then $\alpha n+\beta=\beta$, and hence $\beta\ge\alpha n$, for every $n<\omega$. Hence $\beta\ge\alpha\omega$.
Otoh if $\beta\ge\alpha\omega$ then $\beta=\alpha\omega+\gamma$, hence $\alpha+\beta=\beta$.