Your post expresses the same idea about comparison of infinite sets as in my previous post with the difference being that you use Abel summation while I did use Ramanujan's summation and Zeta regularization. To be more precise, I used Faulhaber's formula for summation because it gives the same results as Ramanujan's summation (I do not know a proper term for this summation method):
$$\sum _{x\ge0}^\Re f(x)=-\sum _{n=1}^\infty \frac {f^{(n-1)}(0)}{n!} B_n(1)$$
Now, addressing your concerns, expressed in the comments to the other answer, there is a set of similar summation methods that are mostly compatible with each other. One can see that some of them depend only on the values of the series at the integer points while others involve integrals and derivatives. So, it clearly means that there can be constructed examples where these methods give different results.
My strong conviction, even if I do not have a proof, is that all these methods should give the same results for "well-behaving" functions. By "well-behaving" I mean functions that are equal to their Newton series:
$$f(x) = \sum_{k=0}^\infty \binom{x-a}k \Delta^k f\left (a\right)$$
I call such functions "discrete-analytic", "Newton-analytic" or "Newtonian".
The criterion can also be written in the following form:
$$f(x)=\lim_{n\to\infty}\frac{\sum_{k=0}^{n} \frac{(-1)^k f(k)}{(x-k)k!(n-k)!}}{\sum_{k=0}^{n} \frac{(-1)^k }{(x-k) k!(n-k)!}}$$
Some functions would have this series divergent, but their two-directional expansion would converge:
$$f(x)=\lim_{n\to\infty}\frac{\sum _{k=-n}^n \frac{(-1)^k f(k)}{(x-k) (k+n)! (n-k)!}}{\sum _{k=-n}^n \frac{(-1)^k}{(x-k) (k+n)! (n-k)!}}$$
I would include such functions into "Newtonian" as well.