You can indeed reduce AC to "AC for well-ordered processes" - however, AWC is "AC for well-ordered index sets," which is quite a different (and weaker) thing.
To see the difference, let's follow your idea for a two-element set $X=\{a,b\}$. Our first "stage" is: pick an element $e_0$ from the set $X$. Our second stage is: pick an element $e_1$ from the set $X\setminus\{e_1\}$.
On the face of it this may seem like we've used the (trivial) axiom of choice for two sets. However, we haven't really done that, since our second set depended on our first choice. For AWC you need to set forth ahead of time a well-ordered set of choices you want to make. But you need a bit more than that to prove WOP: basically, you need "well-ordered dependent choice" rather than simply well-ordered choice. And the problem is that while the set of "stages" in your construction is well-orderable, the set of sets you might need to choose from isn't (at least, a priori), and so mere AWC won't help you here.
Incidentally, note that DC - which is basically "dependent choice for $\omega$-many stages" - is indeed strictly stronger than the axiom of countable choice, which is the "index-set-instead-of-stages" version. So there's a pattern here.
Any theory $T$ over which $\mathsf{WO}$ doesn't imply $\mathsf{AC}$ (that is, which is compatible with $\mathsf{WO+\neg AC}$) must have only very weak set formation ("comprehension") axioms: otherwise, the "pick the least element" argument would get $\mathsf{AC}$ from $\mathsf{WO}$. To find systems this weak we have to go below the usual weak set theories like $\mathsf{KP}$ and $\mathsf{Z}$.
At the risk of self-promotion, let me mention a positive result along these lines. There is a particular theory $\mathsf{RCA_0^3}$, of relevance to higher-order reverse mathematics, about which one can prove the following:
It is consistent with $\mathsf{RCA_0^3}$ that there is a well-ordering of $\mathbb{R}$ but there is a family $(A_r)_{r\in\mathbb{R}}$ of nonempty sets of reals without a choice function.
The proof is a bit technical; there's a plausible alternate argument which is much more elegant, but it has a gap I haven't been able to fill. Very briefly (and not optimizing at all), the idea of the proof I do know is as follows.
Over a countable transitive model $M$ of $\mathsf{ZFC}$ (actually that's massive overkill but meh), we force with the poset $\mathbb{P}$ of countable partial injections from $\mathbb{R}$ to $\omega_1$; let $G$ be the generic bijection $\mathbb{R}\rightarrow\omega_1$, and let $\trianglelefteq$ be the well-ordering of $\mathbb{R}$ gotten by "pulling back" the usual ordering on $\omega_1$ along $G$. Note that recovering $G$ from $\trianglelefteq$ is relatively difficult (even determining $G^{-1}(0)$ requires a kind of unbounded search over $\trianglelefteq$); normally in set theory this wouldn't be significant, but per the first paragraph of this answer we actively want to pay attention to rather fine details re: computational power right now.
A model of $\mathsf{RCA_0^3}$ consists of a natural numbers part, a real numbers part, and a functionals part. We produce such a model $N$ as follows. The naturals and reals of $N$ are just those of $M$ itself, which are in turn those of the generic extension $M[\trianglelefteq]$ since $\mathbb{P}$ is countably closed. As to the functionals, we use a kind of "iterated truth-table reducibility." Roughly speaking, a functional $f\in M[\trianglelefteq]$ is in $N$ iff there is in $M$ an algorithm for computing $f(a)$ (for $a$ a real) with the following behavior: given $a$, we first ask about the $\trianglelefteq$-relationships between some countable set of reals $A_{a,0}$; based on the result of those questions, we whip up another countable set of reals $A_{a,1}$ whose $\trianglelefteq$-relationships we ask about; and so on for some finite number of steps. Now for $r\in\mathbb{R}$ let $$U_r=\{s\in\mathbb{R}: r\triangleleft s\}.$$ This sequence is in $N$, but we can show that no choice function for the sequence $(U_r)_{r\in\mathbb{R}}$ can be "computed" in the sense above.
As an aside, note that the whole iterated truth-table reducibility idea is specific to higher computability theory: in the classical setting it just collapses to truth-table reducibility itself, the point being that "$\mathit{finite}^{\mathit{finite}}=\mathit{finite}$" (which breaks down at higher cardinalities of course).
Best Answer
No.
The axiom "If $X$ can be well-ordered then $2^X$ can be well-ordered" implies the axiom of choice in $\sf ZF$. In $\sf ZFA$ or $\sf ZF-Reg$ this is no longer true, though.
To see more details see the first part of Jech, The Axiom of Choice Chapter 9.
One very interesting observation about the fact that in $\sf ZFA$ this statement does not prove the axiom of choice, is that if $\psi(X)$ is a statement in which all the quantifiers are either bounded in $y$ or bounded in $\mathcal P(y)$ (where $y$ is a variable, of course), and in $\sf ZF$ we have that $\forall X\psi(X)$ implies $\sf AC$, then in $\sf ZFA$ we have that $\forall X\psi(X)$ implies that the power set of a well-ordered set is well-orderable. (This is the first problem in the aforementioned chapter 9.)