There are a number of interesting things to say.

The answer to your first question is yes. Suppose that $M$ is a countable transitive model of set theory and we have a forcing iteration $P_\omega$ in $M$ of length $\omega$, forcing with, say, Cohen forcing $Q_n$ at stage $n$. Let $z$ be any real that cannot be added by forcing over $M$, such as a real that codes all the ordinals of $M$. This real cannot exist in any extension of $M$ to a model of ZFC with the same ordinals. Now, suppose that $G$ is any $M$-generic filter for the iteration, with $G_n$ being the stage $n$ generic filter. Let $H_n$ be the filter that results from $G_n$ by changing the first bit so as to agree with $z(n)$. That is, we change a single bit at each stage. The resulting sequence $\langle H_n | n\in\omega\rangle$ will be generic at every stage, since only finitely many bits are changed by a given stage, but the whole sequence computes $z$, which cannot be added by forcing.

Second, a similar phenomenon occurs even just with 2-step product forcing:

Theorem. If $M$ is a countable transitive model of ZFC, then there are two $M$-generic Cohen reals $c$ and $d$ such that $M[c]$ and $M[d]$ have no common extension to model of ZFC with the same ordinals.

The proof is to build $c$ and $d$ in stages. Fix a real $z$ which cannot exist in any extension of $M$ with the same ordinals, and enumerate the dense sets of $M$ by $D_0, D_1$ and so on. Build $c$ and $d$ in zig-zag fashion: first provide $c_0$ meeting $D_0$, and $d_0$ all $0$s of the same length as $c_0$, followed by the first digit of $z$. Now extend $d_0$ to $d_1$ meeting the dense set, adding all $0$s to $c_0$ making $c_1$ of the same length, and adding one more bit of $z$. And so on. The point is that we ensure that each of $c$ and $d$ is $M$-generic, but together, they reveal the coding points of $z$. So no model extending $M$ with the same ordinals can have both $c$ and $d$, for then it would have $z$.

Third, this is essentially the only kind of obstacle, for there is a positive result here. The following theorem is proved in a current paper I am writing with Gunter Fuchs and Joans Reitz on the topic of set-theoretic geology:

Theorem. If $M$ is a countable transitive model of set theory, and $M[G_n]$ is a sequence of generic extensions of $M$ by forcing $G_n\subset P_n\in M$ of bounded size in $M$, such that the extensions are finitely amalgamble, in the sense that any finitely many of the $M[G_n]$ have a common forcing extension $M[H]$, then there is a single forcing extension $M[H]$ containing all $M[G_n]$.

I'll try to post a proof sketch later, but the main idea is to perform a very large combination of forcing, and then perform surgey so as to replace certain coordinate generics with $G_n$, in such a way that the resulting extension can see only finite fragments of the sequence $\langle G_n | n\lt\omega\rangle$, without being able to construct the whole sequence.

A special case of this theorem answers your second question, in a sense, for if we have a sequence of extensions $M\subset M[G_0]\subset M[G_1]\subset\cdots$, then these extensions are finitely amalgamable, and so there is indeed a common extension $M[H]$ containing every $M[G_n]$. This extension, however, is not an $\omega$-iteration of the forcing notions in your iteration, and in general we cannot expect that the sequence $\langle G_n | n\lt\omega\rangle$ is in $M[H]$, for the reasons described above.

There are a number of interesting things to say.

The answer to your first question is yes. Suppose that $M$ is a countable transitive model of set theory and we have a forcing iteration $P_\omega$ in $M$ of length $\omega$, forcing with, say, Cohen forcing $Q_n$ at stage $n$. Let $z$ be any real that cannot be added by forcing over $M$, such as a real that codes all the ordinals of $M$. This real cannot exist in any extension of $M$ to a model of ZFC with the same ordinals. Now, suppose that $G$ is any $M$-generic filter for the iteration, with $G_n$ being the stage $n$ generic filter. Let $H_n$ be the filter that results from $G_n$ by changing the first bit so as to agree with $z(n)$. That is, we change a single bit at each stage. The resulting sequence $\langle H_n | n\in\omega\rangle$ will be generic at every stage, since only finitely many bits are changed by a given stage, but the whole sequence computes $z$, which cannot be added by forcing.

Second, a similar phenomenon occurs even just with 2-step product forcing:

Theorem. If $M$ is a countable transitive model of ZFC, then there are two $M$-generic Cohen reals $c$ and $d$ such that $M[c]$ and $M[d]$ have no common extension to model of ZFC with the same ordinals.

The proof is to build $c$ and $d$ in stages. Fix a real $z$ which cannot exist in any extension of $M$ with the same ordinals, and enumerate the dense sets of $M$ by $D_0, D_1$ and so on. Build $c$ and $d$ in zig-zag fashion: first provide $c_0$ meeting $D_0$, and $d_0$ all $0$s of the same length as $c_0$, followed by the first digit of $z$. Now extend $d_0$ to $d_1$ meeting the dense set, adding all $0$s to $c_0$ making $c_1$ of the same length, and adding one more bit of $z$. And so on. The point is that we ensure that each of $c$ and $d$ is $M$-generic, but together, they reveal the coding points of $z$. So no model extending $M$ with the same ordinals can have both $c$ and $d$, for then it would have $z$.

Third, this is essentially the only kind of obstacle, for there is a positive result here. The following theorem is proved in a paper with G. Fuchs, myself and J. Reitz on the topic of set-theoretic geology:

Theorem. If $M$ is a countable transitive model of set theory, and $M[G_n]$ is a sequence of generic extensions of $M$ by forcing $G_n\subset P_n\in M$ of bounded size in $M$, such that the extensions are finitely amalgamble, in the sense that any finitely many of the $M[G_n]$ have a common forcing extension $M[H]$, then there is a single forcing extension $M[H]$ containing all $M[G_n]$.

I'll try to post a proof sketch later, but the main idea is to perform a very large combination of forcing, and then perform surgey so as to replace certain coordinate generics with $G_n$, in such a way that the resulting extension can see only finite fragments of the sequence $\langle G_n | n\lt\omega\rangle$, without being able to construct the whole sequence.

A special case of this theorem answers your second question, in a sense, for if we have a sequence of extensions $M\subset M[G_0]\subset M[G_1]\subset\cdots$, then these extensions are finitely amalgamable, and so there is indeed a common extension $M[H]$ containing every $M[G_n]$. This extension, however, is not an $\omega$-iteration of the forcing notions in your iteration, and in general we cannot expect that the sequence $\langle G_n | n\lt\omega\rangle$ is in $M[H]$, for the reasons described above.