Clearly $(a,b,c)\mid (ab+c,bc+a,ca+b)$, so we prove the opposite direction. Let $d=(ab+c,bc+a,ca+b)$. We get $d\mid (a+1)(b+c),(b+1)(c+a),(c+1)(a+b)$. We first show $d\mid (a+b),(b+c),(c+a)$. To that end, assume $d$ has a prime divisor $p$ such that $p\mid a+1$. Let $p\mid b+1$. If $p\mid c+1$ then $(a+1,b+1,c+1)\ge p>1$, a contradiction. If $p\nmid c+1$ then $p\mid a+b$. Using $a\equiv b\equiv -1\pmod{p}$, we get $p=2$. But then $a,b$ are both odd so $c$ must be even. This however is contradictory as $p\nmid ab+c$. So, if $p\mid a+1$ then $p\nmid b+1,c+1$ hence $p\mid a+b$ and $p\mid a+c$. But then $p\mid (a+1,b-1,c-1)$, a contradiction again. So, $p\nmid a+1,b+1,c+1$, and thus $d\mid a+b,b+c,c+a$. Next, $d\mid ab+c\implies d\mid ab-b=a(b-1)$. Similarly, $d\mid b(c-1)$ and $d\mid c(a-1)$. Assume again there is a $p\mid d$ prime so that $p\mid a-1$. Then $(a,a-1)=1$ with $p\mid d\mid a(b-1)$ yields $d\mid b-1$. Similarly, $d\mid c-1$, a contradiction with $(a^2-1,b^2-1,c^2-1)=1$. So, no such prime $p$ exists. As a result, $d\mid a,b,c$ so that $d\mid (a,b,c)$ and therefore $(ab+c,bc+a,ca+b)\mid (a,b,c)$.