No, there do not exist such selective $\mathcal U$ and bijection $\varphi$. Since selectivity is preserved by bijections and since the Fréchet filter $\mathcal N$ is included in $\mathcal F_I\cap\mathcal U$, it suffices to show that no selective ultrafilter $\mathcal U$ on $\omega\times\omega$ includes $\mathcal N\otimes\mathcal N$.

Suppose, toward a contradiction, that we had a selective ultrafilter $\mathcal U\supseteq\mathcal N\otimes\mathcal N$. Partition $\omega\times\omega$ into the columns $\{n\}\times\omega$. Such a column cannot be in $\mathcal U$ because its complement $(\omega-\{n\})\times\omega$ is in $\mathcal N\otimes\mathcal N$ and therefore in $\mathcal U$. So, by selectivity, $\mathcal U$ contains a set $A$ that meets each column at most once. But the complement $(\omega\times\omega)-A$ of this $A$ is in $\mathcal N\otimes\mathcal N$, because it contains a cofinite (in fact at least co-singleton) part of every column. So $(\omega\times\omega)-A\in\mathcal N\otimes\mathcal N\subseteq\mathcal U$, which, together with $A\in\mathcal U$, contradicts the fact that $\mathcal U$ is a filter.

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