@Samuele, just to make sure I understand your argument, are you using implicitly the fact that the group of analytic automorphisms of $\mathbf{C}$ is equal to $z\mapsto az+b$ for $a\neq 0$ so that you know that over a small open set of the base you can normalize in a "coherent way" so that the tangent vector $0$ maps to $0$?

so I just reedited my question.

I see, so missed this subtle point, thanks @ulrich for the explanation

So @Samuele, how do you prove your last statement: "But if $L$ and $C\times\mathbf{C}$ are algebraically isomorphic as varieties, then they are isomorphic as algebraic line bundles".

Hi @Samuele, so your example seems to work. So there must be something wrong in my reasoning...

So what is wrong in the following sketch: By Grauert-Remmert one may extend f to a holomorphic function $\bar{f}:\bar{X}\rightarrow \bar{Y}$ where the bars denote compactifications. Now look at the analytic coherent (algebra)-sheaf $\bar{f}_*O_{\bar{X}}^{an}$. By GAGA it comes from an (unique) algebraic (algebra)-coherent sheaf on $\bar{Y}$. Therefore $X$ and $f$ are algebraic. Isn't?

@Ulrich, Could you give me such an example?

Ok Ulrich, I got it. I was too quick with your key remark: The key point is that $H^{p,q}$ as defined above is $0$ if $p+q>n$ but not if $p+q<n$. Thanks a lot for clearing out my confusion.

You have $f(F^P H)\subseteq F^p H'$ but I don't quite see how you get from that $f(H^{p,q})\subseteq H'^{p,q}$. In fact my counterexample shows that it is wrong, isn't?

But Ulrich I'm confused, you only seem to use the fact that $n\neq n'$ and moreover what about my counterexample