Notation: $\text{HCF}$ is denoted below as $\text{gcd}$.
Assume you have two fractions $\frac{a}{b},\frac{c}{d}$ reduced to lowest
terms. Let
$$\begin{eqnarray*}
a &=&\underset{i}{\prod } p_{i}^{e_{i}(a)},\qquad b=\underset{i}{\prod } p_{i}^{e_{i}(b)}, \\
c &=&\underset{i}{\prod } p_{i}^{e_{i}(c)},\qquad d=\underset{i}{\prod } p_{i}^{e_{i}(d)}.
\end{eqnarray*}$$
be the prime factorizations of the integers $a,b,c$ and $d$. Then
$$\frac{\underset{i}{\prod }\ p_{i}^{\max \left( e_{i}(a),e_{i}(c)\right) }}{%
\prod_{i}\ p_{i}^{\min \left( e_{i}(b),e_{i}(d)\right) }}$$
is a fraction which is a common multiple of $\frac{a}{b},\frac{c}{d}$. It is
the least one because by the properties of the $\text{lcm}$ and $\gcd $ of
two integers, $\prod_{i}\ p_{i}^{\max \left( e_{i}(a),e_{i}(c)\right) }$ is the
least common multiple of the numerators and $\prod_{i}\ p_{i}^{\min \left(
e_{i}(b),e_{i}(d)\right) }$ is the greatest common divisor of the
denominators. Hence
$$\begin{eqnarray*}
\text{lcm}\left( \frac{a}{b},\frac{c}{d}\right) &=&\text{lcm}\left( \frac{{\prod_{i}\ p_{i}^{e_{i}(a)}}}{\prod_{i}\ p_{i}^{e_{i}(b)}},\frac{%
\prod_{i}\ p_{i}^{e_{i}(c)}}{\prod_{i}\ p_{i}^{e_{i}(d)}}\right)=\frac{\prod_{i}\ p_{i}^{\max \left( e_{i}(a),e_{i}(c)\right) }}{%
\prod_{i}\ p_{i}^{\min \left( e_{i}(b),e_{i}(d)\right) }}=\frac{\text{lcm}(a,c)}{\gcd (b,d)}.\quad(1)
\end{eqnarray*}$$
Similarly
$$\begin{eqnarray*}
\gcd \left( \frac{a}{b},\frac{c}{d}\right) =\gcd \left( \frac{{\prod_{i}\ p_{i}^{e_{i}(a)}}}{\prod_{i}p_{i}^{e_{i}(b)}},\frac{%
\prod_{i\ }p_{i}^{e_{i}(c)}}{\prod_{i}p_{i}^{e_{i}(d)}}\right) =\frac{\prod_{i}\ p_{i}^{\min \left( e_{i}(a),e_{i}(c)\right) }}{%
\prod_{i}\ p_{i}^{\max \left( e_{i}(b),e_{i}(d)\right) }} =\frac{\gcd (a,c)}{\text{lcm}(b,d)}.\quad(2)
\end{eqnarray*}$$
The repeated application of these relations generalizes the result to any
finite number of fractions.
Your proof is correct, but you need not use the decomposition of an integer into prime factors. More elementary arguments work here.
As you said, because $\operatorname{gcd}(a,b)=1$, any common multiple of $a$ and $b$ is in fact a multiple of $ab$. Conversely, any multiple of $ab$ is a common multiple of $a$ and $b$.
Hence, the least common multiple must be $ab$ itself.
NB: to justify that any common multiple of $a$ and $b$ is in fact a multiple of $ab$, we can proceed by using Bézout's theorem. We are given a relation $$au+bv=1$$ where $u,v$ are integers. Now, let $m$ be a common multiple of $a$ and $b$. Let us write $m=aa'=bb'$. Then $$m=aa'=(a^2u+abv)a'=(aa')au+(ab)a'v=(bb')au+(ab)a'v=ab(b'u+a'v)$$
With this, the proof is complete with only elementary arguments.
NB2: As @abc... stated, it is in general true that $ab=\operatorname{gcd}(a,b)\operatorname{lcm}(a,b)$. We can now prove this without using the decomposition into prime factors, thanks to the fact that if $k$ is any integer, then $k\operatorname{gcd}(a,b)=\operatorname{gcd}(ka,kb)$ and $k\operatorname{lcm}(a,b)=\operatorname{lcm}(ka,kb)$.
Indeed, given $a,b$, it follows from this property than $\operatorname{gcd}(\frac{a}{\operatorname{gcd}(a,b)},\frac{b}{\operatorname{gcd}(a,b)})=1$. Applying the proven result, we know that $\operatorname{lcm}(\frac{a}{\operatorname{gcd}(a,b)},\frac{b}{\operatorname{gcd}(a,b)})=\frac{ab}{\operatorname{gcd}(a,b)^2}$. Now, multiply both sides by $\operatorname{gcd}(a,b)^2$ to get the result.
Best Answer
Lemma 1: $\operatorname{lcm}(a, n) + \operatorname{lcm} (n-a, n) = \frac{ an } { \gcd(a, n)} + \frac{ (n-a)n} { \gcd(n-a, n)} = \frac{ n\times n} { \gcd(a,n) }$.
Lemma 2: $\sum \frac{n}{\gcd(a,n)} = \sum_{f \mid n} \frac{n}{f} \times \phi(\frac{n}{f} ) = \sum_{d\mid n} d\phi(d)$,
Proof: consider what happens if $ \gcd(a,n) = f \mid n$. It appears $\phi(\frac{n}{f})$ times on the LHS, and each time it has value of $ \frac{n}{f}$. Now substitute $ d = \frac{n}{f}$, which is also a divisor of $n$.
Now, to your problem, pull out $\operatorname{lcm}(n,n) = n$.
We have $ 2 \sum_{a=1}^{n-1} \operatorname{lcm}(a,n) = \sum \left[\operatorname{lcm}(a,n) + \operatorname{lcm} (n-a, n) \right] = n \sum \frac{n}{\gcd(a,n)} = n \times \sum_{d\mid n} d\phi(d).$
Add back $\operatorname{lcm}(n,n)=n$, and you get the formula in OEIS.