Conjecture that $ \frac{\gcd(a+b,ab)}{\gcd(a,b)} \mid \gcd(a,b)$

Write $\gcd(a,b)=d$, then $a=da',b=db'$ and thus $\frac{\gcd(a+b,ab)}{d}=\gcd(a'+b',a'b'd)$ where $\gcd(a',b')=1$. Notice now that $\gcd(a'+b',a'b')=1$ since $a'(a'+b')-a'b'={a'}^2$ and thus $\gcd(a'+b',a'b')|{a'}^2 \implies \gcd(a'+b',a'b')|a'$ and $\gcd(a'+b',a'b')|a'+b' \implies \gcd(a'+b',a'b')|a',b' \implies \gcd(a'+b',a'b')=1$. This means that $\gcd(a'+b',a'b'd)=\gcd(a'+b',d)$ and thus it divides $d$ by definition. So your conjecture is indeed true.

Taking $D=(a,b)$, then $a=DA$ and $b=DB$, $(ab,a+b)=D(A+B,ABD)$ and $(A,B)=1$. So you want to know if $(ab,a+b)/D$ divides $D$, i.e. $(A+B,ABD)|D$? well, let's see if the prime common divisors between $A+B$ and $ABD$ are divisor of $D$ as well.

If $p$ is a prime common divisor of $A+B$ and $ABD$, by Gauss lemma, $p|A$ or $p|B$ or $p|D$. If $p|A$ or $p|B$ there is a contradiction with the coprimality of $A$ and $B$ $p$ cause if $p|A$ then $p|(A+B)-A=B$. So $p|D$.

https://en.wikipedia.org/wiki/P-adic_order

Let $h = \gcd(a+b, ab)$ and $g = \gcd(a,b).$

For each prime factor $p$ including $2,$ two cases:

(I) $$ a = p^k u, b = p^j v $$ with $k > j$ and $u,v \neq 0 \pmod p.$ Then $p$-adic valuation $\nu_p(g) = j.$ Next $\nu_p(ab) = k+j$ while $\nu_p(a+b) = j.$ Put together, $\nu_p(g) = \nu_p(h).$

(II) $$ a = p^k u, b = p^k v $$ with $u,v \neq 0 \pmod p.$ Then $p$-adic valuation $\nu_p(g) = k.$ Next $\nu_p(ab) = 2k$ while $\nu_p(a+b) \geq k.$ Then $$ k \leq \nu_p(h) \leq 2k $$

Put together, $\nu_p(h) \leq 2\nu_p(g).$

In either case, combining all primes, $$ h | g^2 $$

Oh, note that we do have $g | h$ and can write $$ \frac{h}{g} \; | \; g $$