In the definition of a sheaf, the identity axiom is usually stated:
If $g,h\in \mathcal{F}(U)$ and $\{U_i\}_{i\in I}$ is an open cover of $U$ and $\rho^U_{U_i}g=\rho^U_{U_i}h$ for each $i\in I$, then $g=h$.
It seems by writing $\rho^U_{U_i}$ they actually mean $U=\bigcup_{i\in I}U_i$, else we may have for some $i\in I$ that $U_i\cap U\ne\emptyset$ in which case the restriction map doesn't exist. We really mean $U=\bigcup_{i\in I}U_i$ in this axiom right?
Edit: this answer is based on a misreading of the question.
Even if they mean $\bigcup U_i=U$, that doesn't solve the problem, as one of the $U_i$ may be empty. Since the empty set of a topological space is, by convention, open, a presheaf must "know" what are the functions on the empty set, and how to restrict any function to the empty set.
But this is easy. There is one function on the empty set: the empty function. Every function restricts to it. So it's not correct that $\rho^U_{U_i}$ doesn't exist if $U_i$ is empty or disjoint from $U$. It exists, it just restricts everything to one map: the empty map.
It's not required for presheaves restrict everything to a unique element on the empty set, but in fact you can show using the identity axiom that $\mathcal{F}(\varnothing)=*$ for any sheaf $\mathcal{F}.$