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 https://shelah.logic.at/papers/466/ $\operatorname{ZF} + \operatorname{DC}(\omega_1)$ does not prove that there is an injection from $[\mathbb R]^\omega$ to $\mathbb R^\omega$.