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.