I wish to show that if $a+b|ab, a+c|ac, b+c|bc$ then there is a prime which divides all of $a,b,c$, where $a,b,c$ are positive integers.
First, I rule out the case that any pair are coprime: in this case, $(a,b)=(ab,a+b)=1$, but now since $a+b|ab$, we must have that $a+b=1$, which forces one of $a$ or $b=0$, contradicting the positivity of $a,b$. This argument works for any pair, so we've ruled out pairwise co-primality.
Thus, we suppose that $(a,b)=d_{ab}$, $(a,c)=d_{ac}$, $(b,c)=d_{bc}$ and WLOG suppose that $$1<d_{ab}<d_{ac}<d_{bc}$$ (if two are equal, then there is a prime factor of $a,b,c$ and we're done).
I'm having trouble finishing, though. Where can I go from here?
Suppose there is no common factor of all of $a,b,c$. As noted by the OP, each pair of these numbers does have a non-trivial gcd.
Let $a=ZU$ and $b=ZV$, where $U$ and $V$ are coprime. Then $Z(U+V)$ divides $Z^2UV$ and so $U+V$ divides $Z=$ gcd$(a,b)$.
The gcd of $b$ and $c$ is coprime to $Z$ and so divides $V$. Similarly, the gcd of $a$ and $c$ is coprime to $Z$ and so divides $U$.
Hence gcd$(a,b)$ is larger than both gcd$(b,c)$ and gcd$(c,a)$.
We could have started with a different gcd and obtained a different larger gcd. This contradiction proves that there is a common factor of all of $a,b,c$.