If $a,b\in \mathbb{Z}$ then: $$\gcd(a,b)=1 \iff \gcd(a+b,ab)=1$$
Let $p$ be a prime number.
Let $\gcd(a,b)=1$, and $p | a+b,p|ab$. $p|ab \implies p|a \ \text{or}\ p|b$. WLOG let $p|a$, then $p|a+b$ and $p|a$ implies $p|b$. But we have $\gcd(a,b)=1$ a contradiction. Thus We must have $gcd(a+b,ab)=1$.
Conversely let $\gcd(a+b,ab)=1$, $p|a$ and $p|b$ then $p|a+b$ and $p|ab$ a contradiction again. Thus we must have $\gcd(a,b)=1$.
Is the proof correct?
On
You basic reasoning is okay, but as written it is logically sloppy. Here is what it should be like:
Take any integers $a,b$.
If $\gcd(a,b) = 1$: For every prime $p$ we cannot have $p \mid a+b , ab$, otherwise from $p \mid ab$ we get $p \mid a$ or $p \mid b$, and so WLOG $p \mid a$, and from $p \mid a,a+b$ we get $p \mid b$ and so $\gcd(a,b) \ne 1$. Therefore $\gcd(a+b,ab) = 1$.
If $\gcd(a+b,ab) = 1$: For every prime $p$ we cannot have $p \mid a,b$, otherwise we get $p \mid a+b,ab$ and so $\gcd(a+b,ab) \ne 1$. Therefore $\gcd(a,b) = 1$.
Note the key difference between your attempt and my version: The prime $p$ is separately quantified in each case. You cannot use the same prime $p$ in both the forward and reverse direction, because you want to insert the conclusion in each case, namely the statement beginning with "therefore" in my version, which requires the $p$ to be quantified over all primes.
Incidentally, there is another approach:
$\gcd(a+b,ab) = \gcd(a+b,ab-b(a+b)) = \gcd(a+b,b^2)$
$\gcd(a,b) = \gcd(a+b,b) \mid \gcd(a+b,b^2)$. Thus if $\gcd(a+b,ab) = 1$ then $\gcd(a,b) = 1$.
$\gcd(a+b,b^2) \mid \gcd(a+b,b)^2 = \gcd(a,b)^2$. Thus if $\gcd(a,b) = 1$ then $\gcd(a+b,ab) = 1$.
Your proof is perfectly correct. There is nothing to add.
Your proof works also in this implication: Let $m,n\in \mathbb{Z}$, then
$$ \gcd(ma+nb,ab)=1\implies \gcd(a,b)=1$$
and as Heterbij pointed out, the other implication doesn't work.