Disjoint union, cartesian product, function space and power set respect equinumerosity, and for ordinals $\alpha, \beta$ the sets $\alpha \sqcup \beta$ and $\alpha \times \beta$ carry explicit well-orders, so their cardinalities exist in ZF

Fix bijections $f : A \to A'$ and $g : B \to B'$; these exist by [L5], and everything below is built from them, so nothing is chosen beyond one bijection for each of the two hypotheses.

The map $\sigma : A \sqcup B \to A' \sqcup B'$ with $\sigma(0,a) = (0, f(a))$ and $\sigma(1,b) = (1, g(b))$ has the two-sided inverse built the same way from $f^{-1}$ and $g^{-1}$, hence is a bijection.

The map $\pi : A \times B \to A' \times B'$, $\pi(a,b) = (f(a), g(b))$, has the two-sided inverse $(a',b') \mapsto (f^{-1}(a'), g^{-1}(b'))$, hence is a bijection.

The map $\Phi : {}^{B}A \to {}^{B'}A'$, $\Phi(h) = f \circ h \circ g^{-1}$, lands in ${}^{B'}A'$ and has the two-sided inverse $\Psi(h') = f^{-1} \circ h' \circ g$, since $\Psi(\Phi(h)) = f^{-1} \circ f \circ h \circ g^{-1} \circ g = h$ and symmetrically; so it is a bijection and claim (a) holds.

Claim (b): if $f : A \to B$ is a bijection then $S \mapsto f[S]$ maps $\mathcal{P}(A)$ to $\mathcal{P}(B)$ with two-sided inverse $T \mapsto f^{-1}[T]$, hence is a bijection.

On $\alpha \sqcup \beta$ define $(i,\xi) \prec (j,\eta)$ to hold when $i \in j$, or $i = j$ and $\xi \in \eta$; this is irreflexive, transitive and trichotomous by [L6] and [L3], and a nonempty $S \subseteq \alpha \sqcup \beta$ has a $\prec$-least element, namely $(0,\xi_0)$ with $\xi_0$ the $\in$-least $\xi$ having $(0,\xi) \in S$ when such a $\xi$ exists, and $(1,\eta_0)$ with $\eta_0$ the $\in$-least such $\eta$ otherwise.

On $\alpha \times \beta$ define $(\xi,\eta) \lhd (\xi',\eta')$ to hold when $\xi \in \xi'$, or $\xi = \xi'$ and $\eta \in \eta'$; the same three properties hold by [L3] and [L6], and a nonempty $S \subseteq \alpha \times \beta$ has $\lhd$-least element $(\xi_0, \eta_0)$ where $\xi_0$ is the $\in$-least first coordinate occurring in $S$ and $\eta_0$ is the $\in$-least $\eta$ with $(\xi_0,\eta) \in S$; both are least elements of nonempty sets of ordinals, so neither is chosen.

Assuming the Axiom of Choice, ${}^{\beta}\alpha$ carries a well-order by [L7] and therefore has a cardinality by [L1]; this is claim (d), and no step above supplies such a well-order in ZF.

By [L1] applied to the well-orders of steps 1.6 and 1.7, each of $\alpha \sqcup \beta$ and $\alpha \times \beta$ is equinumerous with an ordinal and so has a cardinality in ZF, which is claim (c).

Together: $\sqcup$, $\times$, the function space and the power set all respect $\approx$, the first two have ZF cardinalities on ordinal arguments, and the function space is given one by the Axiom of Choice.