Let me handle just the set-forcing case, since I think the class forcing case is problematic.

Suppose that $\mathbb{P}$
is $\lambda^+$-c.c., and $\mathbb{Q}$ is $\leq\lambda$-closed. If
$G\times H\subset\mathbb{P}\times\mathbb{Q}$ is $V$-generic, then
the claim is that $P(\lambda)^{V[G][H]}=P(\lambda)^{V[G]}$.

First, I claim that $\mathbb{P}$ remains $\lambda^+$-c.c. in
$V[H]$. To see this, if we had a $\mathbb{Q}$-name $\tau$ for an
antichain in $\mathbb{P}$ of size $\lambda^+$, then we may
undertake in $V$ the construction of what is called a
*psuedo*-generic, namely, we build a decreasing sequence of length
$\lambda^+$ of conditions in $\mathbb{Q}$ to decide the next
element of $\tau$. The point is that since $\mathbb{Q}$ is
$\leq\lambda$-closed, we may continue through limits of this
construction. Although the resulting filter that we build is not
actually $V$-generic, every two elements that it decides are in
$\tau$ must be incompatible in $\mathbb{P}$, and this violates the
chain condition in $\mathbb{P}$ in $V$.

(Perhaps you object here that since I am undertaking a transfinite
recursion, I am using the replacement axiom. That is true. My
answer to this objection, however, is that I don't really need a
version of replacement that is reaching arbitrarily high in the
universe, but instead I am merely undertaking a transfinite
recursion to build a descending sequence inside $\mathbb{Q}$,
which is a set. So all I need is to know that the set of all
$\leq\lambda^+$-sequences over $\mathbb{Q}$ exists, which does not
require the replacement axiom. That is, my uses of replacement are
sufficiently bounded, which seems to keep you safe. This reply doesn't work when the forcing is
proper class forcing, since then it seems one is really using replacement.)

Continuing the argument, suppose that $A\subset\lambda$ in
$V[G][H]=V[H][G]$. Thus, $A$ has a $\mathbb{P}$-name in $V[H]$.
Since $\mathbb{P}$ is $\lambda^+$-c.c. in $V[H]$, we may find such
a nice name $\dot A$ having size $\lambda$. Since $\mathbb{Q}$ is
$\leq\lambda$-closed in $V$, it follows that $\dot A\in V$. Thus,
$A={\dot A}_G\in V[G]$, as desired.

The argument generalizes to show that any $\lambda$-sequence in $V[G][H]$ from a set in $V$ is in already $V[G]$. All that is needed for this is to know that if one has $\lambda^+$-c.c. forcing $\sigma$ is the name of a $\lambda$-sequence of elements of a set $B$, then there is a nice name, consisting essentially of a $\lambda$-sequence of $B$-labeled antichains. One argues similarly that such a name cannot have been added by the $\mathbb{Q}$-forcing, and so the name is already in $V$, putting the sequence into $V[G]$, just as with the subsets of $\lambda$. Since everything is bounded here nicely, I don't see that one is using replacement even for the general case of $\lambda$-sequences (although of course without replacement there simply are fewer such sequences that one might have in ZFC).

lessthan $\delta$ have lower bounds. In order to avoid this misinterpretation, I always write either $\leq\delta$-closed or $\lt\delta$-closed. $\endgroup$ – Joel David Hamkins Jan 8 '13 at 14:23