Can one, for an infinite set A in ZFC, use forcing to add so many generic subsets of A as to make the collection of all subsets of A a proper class? Consider now a model $M$ of ZFC and use $Add( , )$ to make the collection of all subsets of any given infinite set in the forcing extension $M$[$G$] a proper class. This forcing extension $M$[$G$] would certainly not be a model of ZFC (the Powerset axiom would fail) but what axioms of ZFC would remain? Also, if replacement would remain and one 'replaces' Replacement with Collection in $M$[$G$], how would that effect $M$[$G$]? 



A question regarding forcing extensions
$\newcommand\Ord{\text{Ord}}
\newcommand\Add{\text{Add}}$Yes, one may undertake such kind of class forcing constructions.
For example, we can force with $\Add(\omega,\Ord)$ to add $\Ord$
many Cohen reals. As you point out, we won't have ZFC in the
resulting forcing extension, but we will get a sensible model of
some theory. The resulting forcing extension will be a model of set theory without power set, $\text{ZFC}^-$, in which the power set of $\omega$ is a proper class. 


Let's be a little more specific. To make things easy to understand
at first, lets start with a countable transitive model
$M\models\text{ZFC}$, and consider forcing with
$\Add(\omega,\Ord)$ from the point of view of $M$. Since this
model is countable, we may find a filter $G$ in this partial order
such that $G$ meets any dense subclass $D\subset\Add(\omega,\Ord)$
that is definable in $M$ using parameters. (There are only
countably many such $D$ and so we can meet them one by one in
order to construct $G$.) Now, we build $M[G]$ by using names in
$M$ and interpreting them by $G$.


Note that the forcing satisfies the countable chain condition by
the usual $\Delta$-system argument. This in effect gives us access
to the Boolean completion of this partial order, from the
perspective of $M$, and we can recursively define the Boolean
value $[\![\varphi(\tau)]\!]$ of any assertion in the forcing
language. We'll have the usual fullness and mixing lemmas for
these Boolean values, and using genericity we'll get that
$M[G]\models\varphi(\tau\_G)$ just in case
$[\![\varphi(\tau)]\!]\in G$. (But in fact, we can use the Boolean algebra to form the Boolean ultrapower of this forcing inside any model of ZFC, not just the countable transitive models.)


Thus, the forcing relation will be definable in $M$ just as for
set forcing. Once you have the definability of the forcing
relation, together with fullness and mixing, then you'll get all
the usual ZFC axioms holding in $M[G]$ for this forcing, including
collection, replacement, separation, choice and so on, except for
the power set axiom. As you mention, we obviously cannot expect
get the power set axiom, since we'll have too many reals in
$M[G]$.


So $M[G]$ will be a model of $\text{ZFC}^-$.


One can similarly add subsets to $\omega\_1$ instead of $\omega$ or
to any other fixed cardinal, with a similar effect. 


There are also interesting situations to arise if one should collapse all cardinals to become countable. This forcing is not c.c.c., but still one gets $\text{ZFC}^-$ in the forcing extension, a model where every set is countable and the reals do not form a set. 


