A divisibility identity for a sum

Reduction. Let $\delta=\divisibility(a,b)$, $a=\delta a'$, $b=\delta b'$, $\divisibility(a',b')=1$. Then
$$a+b=\delta(a'+b'),\qquad \operatorname{lcm}(a,b)=\delta a'b'.$$
So
$$\divisibility(a+b,\operatorname{lcm}(a,b))=\delta\cdot\divisibility(a'+b',a'b').$$

Coprimality. Since $\divisibility(a',b')=1$, any prime dividing $a'b'$ divides exactly one of $a',b'$, hence cannot divide $a'+b'$. Therefore $\divisibility(a'+b',a'b')=1$.

Conclusion. $\divisibility(a+b,\operatorname{lcm}(a,b))=\delta=\divisibility(a,b)$.