Btw @above $p = 3$ seems to be failing; what follows is a solution for $p \geq 5$.

Cute problem. Here's a different solution which provides an explicit construction Since $\prod_{i=1}^{p-1} i \equiv -1 \pmod{p}$ and $\sum_{i=1}^{p-1} i \equiv 1 \pmod{p}$, it suffices to show that there exists $S \subseteq \{1,\ldots,p-1\}$ such that $$\sum_{i \in S} i \equiv -\frac{1}{\prod_{i \in S} i} \pmod{p}.$$If $i \equiv 1 \pmod{4}$, then take $S=\{a\}$, where $a^2 \equiv -1 \pmod{p}$, and both sides are equivalent to $a$. Otherwise, since $p \geq 5$, there exists some odd prime $q \mid p-1$. Then pick $a \in \mathbb{F}_p$ such that $\mathrm{ord}_p(a)=q$, and let $$S=\left\{a^{\frac{q-1}{2}},a^{\tfrac{q-3}{2}},\ldots,a^1,a^{-1},a^{-3},\ldots,a^{-\frac{q-1}{2}}\right\},$$so then $$\prod_{i \in S} \equiv 1 \pmod{p} \text{ and } \sum_{i \in S}i \equiv -1 \pmod{p} \iff a^{\frac{q-1}{2}}\left(a^{\frac{q-1}{2}}+\cdots+a^{-\frac{q-1}{2}}\right)=a^{q-1}+\cdots+1 \equiv 0 \pmod{p},$$which is certainly true, so we're done. $\blacksquare$