It seems to me that this strange situation can indeed happen.

Start with a countable transitive model of set theory $W$,
satisfying CH, and let $G$ be $W$-generic for the forcing
$\newcommand\Add{\text{Add}}\Add(\omega,\omega_1)^W$. So the model
$M=W[G]$ continues to satisfy CH. Now, let $g$ be $W[G]$-generic
for the collapse of $\omega_2^W$ to $\omega_1^W$, using the
collapse forcing as it is defined in $W$, and temporarily consider
$W[G][g]$. Notice that the collapse forcing was countably closed
in $W$, and so it remains at least countably distributive in
$W[G]$. So we have added no new countable sequences of ordinals in
going from $W[G]$ to $W[G][g]$. Inside $W[G][g]$, the ordinals
$\omega_1^W$ and $\omega_2^W$ are now in bijection, and so there
is an isomorphism $\pi:\Add(\omega,\omega_1)^W\cong
\Add(\omega,\omega_2)^W$ between these two forcing notions as posets. Furthermore, since by design
the bijection of the ordinals has the property that every
countable piece of it is in $W$, the same property is true for
this isomorphism $\pi$. Let
$H=\pi[G]\subset\Add(\omega,\omega_2)^W$ be the isomorphic copy of
$G$ induced by the isomorphism $\pi$. I claim that $H$ is
$W$-generic for $\Add(\omega,\omega_2)^W$, because if $A$ is any
maximal antichain in this forcing in $W$, then by the c.c.c. it
follows that $A$ is countable, and so
$\pi^{-1}A\subset\Add(\omega,\omega_1)^W$ is a maximal antichain
in $W$ for the first forcing, since this much of $\pi$ is in $W$.
And so, since $G$ must meet $\pi^{-1}A$, it follows that $H$ meets
$A$; so $H$ is $W$-generic, and we may let $N=W[H]$. So $N$ is a
model of $\neg\text{CH}$, since we've added $\omega_2$ many Cohen
reals. Both $M$ and $N$ are c.c.c. extensions of $W$, and so they
have the same cardinals and cofinalities.

Let's now argue that they have the same reals. First off, every
real of $N=W[H]$ is certainly in $W[G][g]$, where $H$ is
constructed, and the reals of $W[G][g]$ are the same as the reals
of $W[G]=M$, since the $g$ forcing is countably distributive. So
every reals of $N$ is in $M$. Conversely, if $x$ is a real in
$M=W[G]$, then $x$ is in $W[G\upharpoonright\alpha]$ for some
$\alpha<\omega_1$. And since
$\pi\upharpoonright\Add(\omega,\alpha)$ is in $W$, it follows that
$x$ is in $W[H\upharpoonright\pi[\alpha]]$, which is contained in
$W[H]=N$. So they have the same reals.

In conclusion, $M$ and $N$ have the same reals, the same
cardinals, the same cofinalities, but one has CH and one does not,
as desired.

It seems to me that one may cast the argument in terms of forcing over $V$; there is no need to go to countable transitive models. Namely, start in $V$, with CH, and then force to add $\omega_1$ many Cohen reals to form $V[G]$. Now collapse $\omega_2$ to $\omega_1$ using the ground model collapse, and in $V[G][g]$ define $H$ as the copy of $G$ induced by that isomorphism. It now follows by the argument above that $V[G]$ and $V[H]$ have the same reals, the same cardinals and cofinalities, but one has CH and the other does not.

And of course, there is nothing special about $\omega_2$ in these arguments, we could have used $\omega_3$ or any other regular cardinal in $W$ just as easily.