If $(a,b)=1$ then prove $(a+b, ab)=1$.

Let $a$ and $b$ be two integers such that $\left(a,b\right) = 1$. Prove that $\left(a+b, ab\right) = 1$.

$(a,b)=1$ means $a$ and $b$ have no prime factors in common

$ab$ is simply the product of factors of $a$ and factors of $b$.

Let's say $k\mid a+b$ where $k$ is some factor of $a$.

Then $ka=a+b$ and $ka-a=b$ and $a(k-l)=b$.

So $a(k-l)=b, \ a\mid a(k-1)$ [$a$ divides the left hand side] therefore $a\mid b$ [the right hand side].

But $(a,b)=1$ so $a$ cannot divide $b$.

We have a similar argument for $b$.

So $a+b$ is not divisible by any factors of $ab$.

Therefore, $(a+b, ab)=1$.

Would this be correct? Am I missing anything?

Suppose that the gcd is not $1$. Then there is a prime $p$ that divides $ab$ and $a+b$.

But then $p$ divides one of $a$ or $b$, say $a$. Since $p$ divides $a+b$, it follows that $p$ divides $b$. This contradicts the fact that $a$ and $b$ are relatively prime.

Your jump from "$k\mid a+b$ with $k\mid a$" to $ka=a+b$ seems to be wrong. Just because $k$ is a factor of $a$ doesn't mean at all that the number of times it divides $a+b$ is exactly $a$. That seems to kill the rest of your argument.

Instead, here is an approach that doesn't mention prime factors at all. It starts from the well-known property that $(a,b)=1$ exactly if there are $p,q\in\mathbb Z$ such that $pa+qb=1$.

Square $pa+qb=1$ to get $$ p^2a^2 + q^2b^2 + 2pqab = 1 $$ If we can show that each of the terms on the left-hand side of this is an integer combination of $a+b$ and $ab$, then the right-hand side is too.

But $2pqab$ is clearly an integer combination of $a+b$ and $ab$ namely $0(a+b)+2pq\cdot ab$.

And $a^2 = a(a+b)-ab$, so $p^2a^2 = p^2a(a+b)-p^2\cdot ab$.

Similarly $q^2b^2 = q^2b(a+b)-q^2\cdot ab$.

Collecting everything, $(p^2a+q^2b)(a+b)+(2pq-p^2-q^2)ab=1$, so $(a+b,ab)=1$.