I am self-studying Horst Herrlich, Axiom of Choice (Lecture Notes in Mathematics, Vol. 1876), and I'm getting quite confused about cardinal arithmetic without AC.
Here (Which sets are well-orderable without Axiom of Choice?) OP writes that $\mathbb R$ is not necessarily well-orderable in ZF. At the same time, it seems to me that $\mathcal P(\omega)$ is well-orderable in ZF (attempt of proof follows). Therefore, either my proof is wrong, either the answer to the original question is "Yes". Any help? Thanks.
Claim ZF $\vdash$ "$\mathcal P(\omega)$ is well-orderable"
Proof We know that $2=\{0,1\}$ is a cardinal number, therefore it is well-ordered w.r.t. "$\in$" (of course this could be manually checked too). Same holds for $\omega$. (I'll use "$<$" instead of "$\in$"). Let's consider $2^\omega = \{f \mid f : \omega \to 2\}$. The relation defined by $$f \prec g \iff \exists n \, [f(n) < g(n) \wedge \forall i\!\!<\!\!n \;\; f(i)=g(i)]$$ is a well-order on $2^\omega$ (and this is true in ZF). To conclude, just observe that ZF $\vdash 2^\omega \approx \mathcal P(\omega)$.
Update: The proof is wrong, since the lexicographic order on $2^\omega$ is clearly not a well-order, as pointed out in Asaf Karagila's answer. Asaf claims also that ZF $\not\vdash$ "$2^\omega$ is well-orderable". But then, I have another question: in the book I cite above, there is the following proof (page 51):
Now: If $2^\omega$ is not necessarily well-orderable, why is the author allowed to use $2^{\aleph_0}$ as a cardinal number? Or, equally, why is he allowed to talk about the cardinality of $\prod_{n \in \mathbb N} B_n$?
Solution to update: It was just a matter of definitions. In the text I've studied these topics, cardinal numbers are defined as ordinals for which there is no bijection with smaller ordinals. Then, given a set $A$, $|A|$ is a cardinal number, therefore it is defined iff $A$ is well-orderable. On the contrary, in this book $|A|$ is intended as the equivalence class of $A$ under the equivalence relation $A \sim B \iff $ there exist a bijection $A \to B$. This way, $|A|$ is always well-defined.
You don't need the axiom of choice to show that $|\Bbb R|=2^{\aleph_0}$. If you observe carefully neither the injection from $\Bbb R$ into $\mathcal P(\omega)$ defined by enumerating the rationals and using Dedekind cuts, nor the injection from $\mathcal P(\omega)$ into $\Bbb R$ defined by creating the Cantor set use the axiom of choice.
Since the proof of the Cantor-Bernstein theorem does not rely on the axiom of choice either, this shows that $\sf ZF\vdash|\Bbb R|=|\mathcal P(\omega)|$.
As for your proof that $2^\omega$ is a well-orderable set, this relation is not well-founded (so it cannot be a well-order). Let $f_n$ be the function which is constant $0$ until $n$ and then $1$ from $n$ onwards. Then $f_n\prec f_k$ if and only if $n>k$ which means this is a decreasing sequence.
To your edit, note that cardinal exponentiation is still well-defined. Just as a product of two cardinals need not be equal to the infinite sum of one by an index of the other, the exponentiation need not be the product of one by itself over the index of the other.
This just gives more weight to the old saying that multiplication is not repeated addition (and exponentiation is not repeated multiplication).
We define the cardinal arithmetic directly: $|A|+|B|$ is the cardinality of $|A\cup B|$ where $A\cap B=\varnothing$; $|A|\cdot|B|=|A\times B|$; and $|A|^{|B|}=|A^B|$, namely the cardinality of all functions from $B$ into $A$.
And note that $\prod_{i\in I}X_i$ is a particular set, so it has a well-defined cardinality. Whether or not it is equal to $\prod_{i\in I}Y_i$ under the assumption that $|X_i|=|Y_i|$ for all $i\in I$ is a whole other story. But particular products have particular cardinalities.