Does a finite sum of distinct prime reciprocals always give an irreducible fraction?

For distinct primes $p_1,\dots,p_n$ the fraction $S=\frac{1}{p_1}+\cdots+\frac{1}{p_n}$ is always irreducible. For if it were reducible, some $p_i$ would "cancel", without loss of generality $p_1$. So we could express $S$ as $\frac{A}{p_2p_3\cdots p_n}$.

Multiply both sides of the equation $$\frac{1}{p_1}+\frac{1}{p_2}+\cdots+\frac{1}{p_n}=\frac{A}{p_2p_3\cdots p_n}$$ by $p_2p_3\cdots p_n$. On the right-hand side we obtain an integer. On the left-hand side we do not.

As you've shown, this comes down to the question of whether $$P=\sum_{k=1}^n p_1p_2\ldots\hat p_k\ldots p_n$$ and $p_1\ldots p_n$ are coprime, where $\hat a$ means that the product excludes $a$.

The only candidates for common prime factors are the $p_i$. But

$$S\equiv p_1p_2\ldots\hat p_i\ldots p_n\pmod {p_i}$$which is nonzero as the primes are distinct and not equal to $p_i$. Hence $p_i\nmid S$ for each $i$.