Let $d=\gcd(a,b,c)$ and let $e=\gcd(bc+a,ca+b,ab+c)$.
Our goal is to show $e=d$.
First a lemma . . .
Lemma:$\;$If $p$ is prime and $p{\,\mid\,}e$, then $p{\,\mid\,}d$.
Proof of the lemma:
Suppose $p$ is prime and $p{\,\mid\,}e$.
From $\gcd(a^2-1,b^2-1,c^2-1)=1$, it follows that at least one of $a^2-1,b^2-1,c^2-1$ is not a multiple of $p$.
Assume $p{\,\not\mid\,}(c^2-1)$.
\begin{align*}
\text{Then}\;\;&
p{\,\mid\,}e
\\[4pt]
\implies\;&
p{\,\mid\,}(bc+a),\;\;p{\,\mid\,}(ca+b)
\\[4pt]
\implies\;&
bc\equiv -a\;(\text{mod}\;p),\;\;ca\equiv -b\;(\text{mod}\;p)
\\[4pt]
\implies\;&
c^2ab\equiv ab\;(\text{mod}\;p)
\\[4pt]
\implies\;&
ab(c^2-1)\equiv 0\;(\text{mod}\;p)
\\[4pt]
\implies\;&
ab\equiv 0\;(\text{mod}\;p)
\\[4pt]
\implies\;&
p{\,\mid\,}ab
\\[4pt]
\implies\;&
p{\,\mid\,}\bigl((ab+c)-ab\bigr)
\\[4pt]
\implies\;&
p{\,\mid\,}c
\\[4pt]
\implies\;&
p{\,\mid\,}\bigl((bc+a)-bc\bigr),\;\;p{\,\mid\,}\bigl((ca+b)-ca\bigr)
\\[4pt]
\implies\;&
p{\,\mid\,}a,\;\;p{\,\mid\,}b
\\[4pt]
\end{align*}
Thus $p{\,\mid\,}a,\;p{\,\mid\,}b,\;p{\,\mid\,}c$, so $p{\,\mid\,}d$.
This completes the proof of the lemma.
Returning to the main proof, write
$$
\left\lbrace
\begin{align*}
a&=da_1\\[4pt]
b&=db_1\\[4pt]
c&=dc_1\\[4pt]
\end{align*}
\right.
$$
where $\gcd(a_1,b_1,c_1)=1$.
Then we get
$$
e=d{\,\cdot\,}\gcd(db_1c_1+a_1,dc_1a_1+b_1,da_1b_1+c_1)
$$
so it remains to show $\gcd(db_1c_1+a_1,dc_1a_1+b_1,da_1b_1+c_1)=1$.
Suppose instead that $\gcd(db_1c_1+a_1,dc_1a_1+b_1,da_1b_1+c_1) > 1$.
Let $p$ be a prime factor of $\gcd(db_1c_1+a_1,dc_1a_1+b_1,da_1b_1+c_1)$.
Then $p{\,\mid\,}e$, so by the lemma, we get $p{\,\mid\,}d$, hence
$$
\left\lbrace
\begin{align*}
&
p{\,\mid\,}\bigl((db_1c_1+a_1)-db_1c_1\bigr)
\\[4pt]
&
p{\,\mid\,}\bigl((dc_1a_1+b_1)-dc_1a_1\bigr)
\\[4pt]
&
p{\,\mid\,}\bigl((da_1b_1+c_1)-da_1b_1\bigr)
\\[4pt]
\end{align*}
\right.
$$
But that gives $p{\,\mid\,}a_1,\;p{\,\mid\,}b_1,\;p{\,\mid\,}c_1$, contrary to $\gcd(a_1,b_1,c_1)=1$.
Hence $\gcd(db_1c_1+a_1,dc_1a_1+b_1,da_1b_1+c_1)=1$, so $e=d$, as was to be shown.
