If you say the other axioms correctly, then the condition on $\operatorname{End}(1)$ is redundant. Indeed, the word "tensor functor" implies that $\omega: 1 \mapsto k$, and the word "faithful" implies that $\operatorname{End}(1) \hookrightarrow \operatorname{End}(\omega(1))$. What you should include that you don't on your list is that $\omega$ be $k$-linear. You should also demand that $\mathcal C$ be a nontrivial category; then you cannot have $\operatorname{End}(1) = 0$, as $\operatorname{End}(1)$ acts on all other homsets via the $1$ action and in particular $\operatorname{id}\in \operatorname{End}(1)$ acts as the identity on all other homsets. With all of this, it follows that $\operatorname{End}(1) = k$.

Conversely, you can see the condition that $\operatorname{End}(1) = k$ as being a "nontriviality" condition. It is necessary only to assure that $\mathcal C \neq 0$. In particular, no group has zero representation theory, as every group has a trivial representation on $k$.

If you believed in "the empty group", then you would not need this restriction: the zero category is the category of representations of the zero ring, which is "the group ring of the empty group".