Proof that if gcd(a, b) = 1, then lcm(a, b) = ab

Question

I am trying to show the following. If $gcd(a,b) = 1$ then $lcm(a, b)=ab$. I am assuming a and b are positive integers.

I have looked at the following questions on here which are using different presumed theorems that I don't have.

first link
second link

But both assume the notion of prime factorisation. I am working out the exercises in the appendix of Pinters - "A book of abstract algebra" before starting the book proper. I only have the following to work with: Bezouts lemma, definitions of gcd, lcm, relatively prime integers and definition of division, multiples etc.

My attempt to prove it is as follows:

Proof: Assume $a$ and $b$ are integers and $gcd(a,b)=1$. Then $a$ and $b$ are relatively prime and it follows that there exists integers $k$ and $l$ such that $1 = ak + bl$. Let $m$ be a common multiple of $a$ and $b$. Then $m=ar$ and $m=bs$ for some integers $r$ and $s$. Multiplying $1=ak+bl$ through by m we have:
$m = akm +blm = ak(bs) + bl(ar) = (ab)(ks + lr)$. As $ks + lr$ is an integer $ab|m$. Clearly $a|ab=m$ and $b|ab=m$

Hence $a|ab$, $b|ab$ and for any multiple of $a$ and $b$, $m$, $ab|m$ hence $ab=lcm(a,b)$ $\square$

Any corrections are welcome.
Again, as I said before I have limited definitions and theorems to work with so I have to stick to basic approaches.
If I were to try to prove this for $a$, $b$ negative or positive how would I proceed? would I need to do four cases to prove it for the various cases where $a$ and $b$ are positive and negative?