Easiest and most complex proof of $\gcd (a,b) \times \operatorname{lcm} (a,b) =ab.$

Let $\gcd(a,b)=d$. Then for some $a_0,b_0$ such that $a_0$ and $b_0$ are relatively prime, we have $a=da_0$ and $b=d b_0$. If we can show that the lcm of $a$ and $b$ is $da_0b_0$, we will be finished.

Certainly $da_0b_0$ is a common multiple of $a$ and $b$. We must show that it is the least common multiple.

Let $m$ be a common multiple of $a$ and $b$. We will show that $da_0b_0$ divides $m$.

Since $m$ is a multiple of $a$, we have $m=ka=ka_0d$ for some $k$. But $b$ divides $m$, so $db_0$ divides $ka_0d$, and therefore $b_0$ divides $ka_0$. Since $a_0$ and $b_0$ are relatively prime, it follows that $b_0$ divides $k$, and we are finished.

First notice that $$ \dfrac{ab}{\gcd(a,b)} = a\dfrac{b}{\gcd(a,b)} = b\dfrac{a}{\gcd(a,b)} $$ is a common multiple of $a$ and $b$. By the minimality of the $\operatorname{lcm}$, $$ \frac{ab}{\gcd(a,b)}\ge\operatorname{lcm}(a,b)\Longrightarrow ab\ge\operatorname{lcm}(a,b)\gcd(a,b)\tag{1} $$ By division, we can write $$ ab = q\operatorname{lcm}(a,b) + r\quad\text{where}\quad0 \le r \lt \operatorname{lcm}(a,b) $$ Because $ab$ and $\operatorname{lcm}(a,b)$ are common multiples of $a$ and $b$, so is $r$. By the minimality of the $\operatorname{lcm}$, $r = 0$. Therefore, $\operatorname{lcm}(a,b)$ divides $ab$. Notice that $$ \frac{ab}{\operatorname{lcm}(a,b)} = \frac{a}{\operatorname{lcm}(a,b)/b} = \frac{b}{\operatorname{lcm}(a,b)/a} $$ is a common divisor of $a$ and $b$. By the maximality of the $\gcd$, $$ \frac{ab}{\operatorname{lcm}(a,b)} \le \gcd(a,b)\Longrightarrow ab\le\operatorname{lcm}(a,b)\gcd(a,b)\tag{2} $$ Combining $(1)$ and $(2)$, we get $$ ab = \operatorname{lcm}(a,b)\gcd(a,b) $$

The following is more general than for the integers, and therefore simpler (but longer than a proof using unique factorisation without proving it; here we start from scrap).

Let $R$ be an integral domain, where $d=\gcd(a,b)$ is defined to mean that $d\mid a,b$ and $d'\mid a,b\implies d'\mid d$ for all $d'\in R$, while $\def\lcm{\operatorname{lcm}}m=\lcm(a,b)$ is defined to mean that $a,b\mid m$ and $a,b\mid m'\implies m\mid m'$ for all $m'\in R$ (in both cases it is not implied that $\gcd(a,b)$ or $\lcm(a,b)$ always exist, and if they do they are only unique up to multiplication by invertible elements; as a consequence in this setting the equality $\gcd(a,b)\times\lcm(a,b)=ab$ can only be asserted up to such multiplication, or for properly chosen values on the left hand side).

Lemma. Let $r\in R\setminus\{0\}$, and put $D_r=\{\, d\in R: d\mid r\,\}$, the set of divisors of $r$. Then $f_r:d\mapsto r/d$ defines an involution of $D_r$ which is an anti-isomorphism for the divisibility relation: for $a,b\in D_r$ one has $a\mid b\iff f(b)\mid f(a)$.

Proof. Since by definition $d f(d)=r$ for all $d\in D_r$ one has $f(d)\in D_r$ and $f(f(d))=d$. Suppose $a,b\in D_r$ satisfy $a\mid b$, so there exists $c\in R$ with $ac=b$, then $r=bf(b)=acf(b)$ so $f(a)=cf(b)$ and $f(b)\mid f(a)$. Conversly if $f(b)\mid f(a)$ applying this result gives $f(f(a))\mid f(f(b))$ which simplifies to $a\mid b$. QED

Proposition. If $ab\neq0$ and $m=\lcm(a,b)$ exists, then $ab/m=\gcd(a,b)$.

Proof. One has $a,b\mid ab$ so $m\mid ab$ by definition of the $\lcm$; therefore $a,b,m\in D_{ab}$. One has $f_{ab}(a)=b$ and $f_{ab}(b)=a$, and since $a,b\mid m$ one has $f_{ab}(m)\mid b,a$ by the lemma. Also if $d'\in R$ satisfies $d'\mid a,b$ then $d\in D_{ab}$ so $b,a\mid f_{ab}(d')$ by the lemma, whence $m\mid f_{ab}(d')$ by definition of the $\lcm$, and once again by the lemma $d'\mid f_{ab}(m)$. Thus $$ab/m=f_{ab}(m)=\gcd(a,b). \qquad\text{QED}$$

Concluding $\gcd(a,b)\times \lcm(a,b)=ab$ needs the precaution that it only holds if $\lcm(a,b)$ exists, and then the left hand side is defined up to invertible factors only, so the equality should be interpreted in this sense. For the case $ab=0$ not covered by the proposition one has $0=\lcm(a,b)$ and $\{a,b\}=\{0,\gcd(a,b)\}$, so the equality holds without any difficulty.

Note that the existence of $\gcd(a,b)$ does not imply the existence of $\lcm(a,b)$ in general.