Since the question has arisen whether the standard
arguments really do work with classes, let me post this
answer giving some fuller details about one comparatively
concrete way to do it. Other methods are also possible.

I shall work in the theory of Goedel-Bernays GB set theory
(without global choice), a general setting for treating
classes, which forms a conservative extension of ZF. This
theory includes as a special case the traditional treatment
of classes as definable collections in ZF, since every
model of ZF, when augmented with its definable classes,
forms a model of GB. Thus, the GB context seems the most
comprehensive way to answer (and if you prefer ZF, then
just imagine that all classes here are definable from
parameters).

Suppose that $A$ and $B$ are classes and that we have class
functions $F:A\to B$ and $G:B\to A$ which are injective, as
in the question. Let $A_0=A-\operatorname{ran}(G)$, which is a class,
since it is definable from $A$ and $G$. Let us say that a
sequence $\langle x_0,x_1,x_2,\ldots\rangle$ is a
*back-and-forth-iteration sequence* if $x_0\in A$ and
$x_{2n+1}=F(x_{2n})$ and $x_{2n+2}=G(x_{2n+1})$ for all natural numbers $n$. Let $A_n$
be the elements $a_{2n}$ that appear at the even
coordinates of a back-and-forth iteration sequence with
$a_0\in A_0$. This notion is definable from $F$ and $G$, and the point is that we have a uniform presentation of the $A_n$ in a single class $\{(n,a)\mid a\in A_n\}$.
Let $A^+=\bigcup_n A_n$, which is a class definable from
$F$ and $G$.

Let $H$ be the function $(F\upharpoonright
A^+)\cup(G^{-1}\upharpoonright A-A^+)$. I claim that this
is the desired bijection between $A$ and $B$. First, it is
clearly a class that is definable from $F$ and $G$. Second,
it is a function from $A$ to $B$. Note that if $a\in A_n$,
then $G(F(a))\in A_{n+1}\subset A^+$, and so $F(a)$ is not
$G^{-1}(a')$ for any $a'\in A-A^+$. Thus, the function $H$
is injective. Secondly, if $b\in B$ and $G(b)\in A_n$, then
it must be that $n\geq 1$ and so $b=F(a')$ for some $a'\in
A_{n-1}\subset A^+$, putting $b\in\operatorname{ran}(H)$. Otherwise,
$G(b)\notin A^+$, and so again $b\in\operatorname{ran}(H)$. So $H$ is a
bijection. QED

Some of the other proofs can also be formalized for
classes, if one simply uses the sequences as I did here for
iterating.