Recall that $\mathcal Z$ is a weak $P$-point if it is not in the closure of any countable, relatively discrete subset of $\omega^*$. A weak $P$-point is never the tensor product of two non-principal ultrafilters.

To see this, suppose $\mathcal U$ and $\mathcal V$ are two non-principal ultrafilters on $\omega$, and let $\mathcal Z = \mathcal U \times \mathcal V$. I will show that $\mathcal Z$ is not a weak $P$-point in the space of non-principal ultrafilters on $\omega \times \omega$. (This implies that $\psi(\mathcal Z)$ is not a weak $P$-point in $\omega^*$ for any bijection $\psi: \omega \times \omega \rightarrow \omega$.)

For each $i \in \omega$, let $$v_i = \{\{i\} \times B \,:\, B \in \mathcal V\},$$ and observe that $v_i$ is a non-principal ultrafilter on $\omega \times \omega$. The set $\{v_i \,:\, i \in \omega\}$ is a countable, relatively discrete subset of $(\omega \times \omega)^*$. (It is relatively discrete because for each $i \in \omega$, $U_i = (\{i\} \times \omega)^*$ is a (cl)open subset of $(\omega \times \omega)^*$ such that $U_i \cap \{v_i \,:\, i \in \omega\} = \{v_i\}$.) Yet we have $\mathcal Z \in \overline{\{v_i \,:\, i \in \omega\}}$. Indeed, it is clear from the definition of $\mathcal Z$ that every neighborhood of $\mathcal Z$ in $(\omega \times \omega)^*$ contains $\mathcal U$-many of the $v_i$. (In what might be more familiar notation, this means $\mathcal Z \,=\, \mathcal U\hspace{-.5mm}-\lim_{i \in \omega} v_i$.) Thus $\mathcal Z$ is not a weak $P$-point.

Finally, let me mention that Kunen famously proved that $\omega^*$ contains weak $P$-points. Thus the above observation shows that some ultrafilters are not representable as tensor products.

Recall that $\mathcal Z$ is a weak $P$-point if it is not in the closure of any countable subset of $\omega^* \setminus \{\mathcal Z\}$. A weak $P$-point is never the tensor product of two non-principal ultrafilters.

To see this, suppose $\mathcal U$ and $\mathcal V$ are two non-principal ultrafilters on $\omega$, and let $\mathcal Z = \mathcal U \times \mathcal V$. I will show that $\mathcal Z$ is not a weak $P$-point in the space of non-principal ultrafilters on $\omega \times \omega$. (This implies that $\psi(\mathcal Z)$ is not a weak $P$-point in $\omega^*$ for any bijection $\psi: \omega \times \omega \rightarrow \omega$.)

For each $i \in \omega$, let $$v_i = \{\{i\} \times B \,:\, B \in \mathcal V\},$$ and observe that $v_i$ is a non-principal ultrafilter on $\omega \times \omega$. The set $\{v_i \,:\, i \in \omega\}$ is a countable, relatively discrete subset of $(\omega \times \omega)^*$. (It is relatively discrete because for each $i \in \omega$, $U_i = (\{i\} \times \omega)^*$ is a (cl)open subset of $(\omega \times \omega)^*$ such that $U_i \cap \{v_i \,:\, i \in \omega\} = \{v_i\}$.) Clearly $\mathcal Z$ is not equal to any of the $v_i$. Yet we have $\mathcal Z \in \overline{\{v_i \,:\, i \in \omega\}}$. Indeed, we see from the definition of $\mathcal Z$ that every neighborhood of $\mathcal Z$ in $(\omega \times \omega)^*$ contains $\mathcal U$-many of the $v_i$. (In what might be more familiar notation, this means $\mathcal Z \,=\, \mathcal U\hspace{-.5mm}-\lim_{i \in \omega} v_i$.) Thus $\mathcal Z$ is not a weak $P$-point.

Finally, let me mention that Kunen famously proved that $\omega^*$ contains weak $P$-points. Thus the above observation shows that some ultrafilters are not representable as tensor products.