If we define a function f from N to a well ordered set E, such that 0 is mapped to the smallest element in E, and the successor (in E) of the f-image of an integer, is the f-image of the successor of said integer.
Intuitically, I'd say f is surjective, but I don't know how to prove it.
You can't prove it, because it's false even if $E$ is countable. Consider $E=\mathbb{N}\cup\{\infty\}$, with the order $$ a\le b \qquad\text{for}\qquad b=\infty\text{ or }(a,b\in\mathbb{N}\text{ and }a\le b) $$ Prove that $E$ is well-ordered by $\le$.
In this case the function you define is not surjective, because $\infty$ is not in the image of $f$, being the successor of no integer.