If $A = \{0\}\times 2^{\omega}$ and $B = (\{0\}\times 2^{\omega})\cup (\{1\}\times \omega_1)$ and there is no injection from $\omega_1$ to $2^{\omega}$,
then there does not exist such a bijection.
Define $f : (\{0\}\times 2^{\omega}) \to ((\{0\}\times 2^{\omega})\cup (\{1\}\times \omega_1))$ by $f(x) = x$. Obviously, $f$ is injective.
Define $\operatorname{pair} : \omega^2 \to \omega$ by $\operatorname{pair}(\langle m,n\rangle) = ((m+n)\cdot (m+n+1))+(2\cdot n)$.
Define $g_1 : 2^\{\omega\} \to \omega_1$ by
$g_1(x) = \begin{cases} \alpha & \text{if } \{\langle m,n\rangle \in \omega^2 : \operatorname{pair}(\langle m,n\rangle) \in x\} \text{ is a well-order of } \omega \text{ with order type } \alpha \\ 0 & \text{else} \end{cases} \quad .$
Define $g : (\{0\}\times 2^{\omega}) \to ((\{0\}\times 2^{\omega})\cup (\{1\}\times \omega_1))$ by
$g(\langle 0,x\rangle) = \begin{cases} \langle 0,\{n\in \omega : (n+1) \in x\}\rangle & \text{if } 0 \notin x \\ \langle 1,g_1(\{n\in \omega : (n+1) \in x\})\rangle & \text{if } 0\in x\end{cases} \quad .$
Since $g_1$ is surjective, $g$ is also surjective. Let $h : ((\{0\}\times 2^{\omega})\cup (\{1\}\times \omega_1)) \to (\{0\}\times 2^{\omega})$ be a function. Define $h_1 : \omega_1 \to 2^{\omega}$ by $h_1(\alpha) = \operatorname{secondentry}(h(\langle 1,\alpha \rangle))$.
$h_1$ is not injective, so $h$ is not injective either.
This works for all functions $h : ((\{0\}\times 2^{\omega})\cup (\{1\}\times \omega_1)) \to (\{0\}\times 2^{\omega})$, so
there does not exist a bijection $h : ((\{0\}\times 2^{\omega})\cup (\{1\}\times \omega_1)) \to (\{0\}\times 2^{\omega})$.
if $A = \{\}$ and $B = \{\}$, then there does exist such a bijection.
Define $h : A\to B$ by $h(x) = x$.
Therefore there does not have to be a single answer, it can depend on $A$ and $B$.
If all sets of reals have Baire property, then there is no injection from $[\mathbb R]^\omega$ to $\mathbb R^\omega$.
Let $h : [\mathbb R]^\omega \to \mathbb R^\omega$ be a function. Define $\leq$ as the lexicographic order of $\mathbb R^\omega$.
By this answer there is no total order of ${\mathbb R}/\sim$. All members of ${\mathbb R}/\sim$ are countable subsets of $\mathbb{R}$.
Define $\leq_h$ on ${\mathbb R}/\sim$ by $x\: \leq_h\: y$ if and only if $h(x) \leq h(y)$. Since $\leq_h$ is not a total order, $h$ is not injective. This works for all functions $h : [\mathbb R]^\omega \to \mathbb R^\omega$, therefore there is no injection $h : [\mathbb R]^\omega \to \mathbb R^\omega$.
By shelah.logic.at/files/446.ps (which I can't figure out how to make markdown link to),
$\operatorname{ZF} + \operatorname{DC}(\omega_1)$ does not prove that there is an injection from $[\mathbb R]^\omega$ to $\mathbb R^\omega$.
Yes, your remarks about incomparability of sets and classes without the axiom of choice are correct.
Yes, in ZF (or in GB), the axiom of choice is equivalent to the assertion that every set injects into every proper class (one must say proper class, since in the usual terminology, every set also counts as a class). The forward direction holds, since every proper class $A$ must have elements of unboundedly many ranks, and so it has arbitrarily large subclasses. Conversely, if a set $a$ injects into the ordinals, then it is well-orderable.
In ZF or GB, that is, without the axiom of choice, one can interpret the notion of proper class as those classes that do not contain all their elements at some stage of the cumulative hierarchy. That is, they must contain elements of ranks unbounded in the ordinals. So the essence of what makes something a proper class is not its size, as such, but rather the fact that at no stage of the cumulative hierarchy did one finish completing the class in the sense of already having all of the elements of the class by that stage.
Meanwhile, one can recover the idea that proper classes are large, even in ZF or GB, by considering quotients of the class, for a class $A$ is a proper class if and only if there is an equivalence relation $\sim$ on $A$ (and one may freely assume that all equivalence classes are sets) such that the class $\text{Ord}$ injects into the quotient $A/\sim$. Indeed, the same equivalence relation $\sim$ works for all classes $A$: let $a\sim b$ if and only if they have the same rank. A class $A$ is a proper class if and only if it has elements from $\text{Ord}$ many distinct ranks.
Best Answer
Ignoring set-theoretic technicalities of formulating the question properly, I see no reason that the usual proof of Schroder-Bernstein wouldn't work.
(Set-theoretic technicalities: In the standard language of set theory, you can't quantify over classes, so you can't quite state this. However, you can prove a metatheorem saying that whenever you exhibit two such injections, you can prove there is also a bijection. Alternatively, you could work in set theory with classes, in which the statement can be made properly and you ought to be able to prove it just like ordinary Schroder-Bernstein. Alternatively, it is a trivial corollary of the "global" axiom of choice (which implies, in particular, that all proper classes have the same size), though this is kind of applying a sledgehammer.)