One easy and insightful way is to use the proof below. It essentially constructs $\rm\:gcd\:$ from $\rm\:lcm\:$ by employing duality between minimal and maximal elements - see the Remark below. This is essentially how the linked Wikipedia proof works, but there the innate duality is obfuscated by the presentation. Below is a proof structured so that this fundamental duality is brought to the fore.
$\rm{\bf Theorem}\quad c\mid a,b\iff c\mid d,\ \ $ for $\rm\ \ d = ab/lcm(a,b).\ $ $\rm\color{#0a0}{Hence}$ $\rm\ d = gcd(a,b)$
$\rm{\bf Proof}\qquad\ \ \, c\mid a,b \iff a,b\mid ab/c \iff lcm(a,b)\mid ab/c \iff c\mid ab/lcm(a,b)$
$\rm\color{#0a0}{Generally}\,$ if $\rm\, c\mid a,b\iff c\mid d\ $ then $\rm\ d = \gcd(a,b)\ $ up to unit factors, i.e. they're associate.
Indeed setting $\rm\:c = d\:$ in direction $(\Leftarrow)$ shows that $\rm\:d\mid a,b,\:$ i.e. $\rm\:d\:$ is a common divisor of $\rm\:a,b.\:$ Conversely, by direction $(\Rightarrow)$ we deduce that $\rm\:d\:$ is divisible by every common divisor $\rm\:c\:$ of $\rm\:a,b,\:$ thus $\rm\:c\mid d\:\Rightarrow\: c\le d,\:$ so $\rm\:d\:$ is a greatest common divisor (both divisibility and magnitude-wise).
Remark $\ $ The proof shows that, in any domain, if $\rm\:lcm(a,b)\:$ exists then $\rm\:gcd(a,b)\:$ exists and $\rm\ gcd(a,b)\,lcm(a,b) = ab\ $ up to unit factors, i.e. they are associate. The innate duality in the proof is clarified by employing the involution $\rm\ x'\! = ab/x\ $ on the divisors of $\rm\:ab.\:$ Let's rewrite the proof using this involution (reflection).
Notice that $\rm\ x\,\mid\, y\:\color{#c00}\iff\: y'\mid x'\,\ $ by $\smash[t]{\,\ \rm\dfrac{y}x = \dfrac{x'}{y'} \ }$ by $\rm\, \ yy' = ab = xx',\ $ so rewriting using this
$\begin{eqnarray}\rm the\ proof\ \ \ c\mid a,b &\iff&\rm b,\,a\mid ab/c &\iff&\rm lcm(b,\,a)\mid ab/c &\iff&\rm c\mid ab/lcm(b,a)\\[.5em]
\rm becomes\ \ \ \ c\mid a,b &\color{#c00}\iff&\rm a',b'\mid c' &\iff&\rm lcm(a',b')\mid c' &\color{#c00}\iff&\rm c\mid lcm(a',b')'\end{eqnarray}$
Now the innate duality is clear: $\rm\ gcd(a,b)\,=\,lcm(a',b')'\ $ by the $\rm\color{#0a0}{above}$ gcd characterization.
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.
Best Answer
That is not correct, because Bézout's identity doesn't say that.
However, it can be corrected. It follows from what you did that$$\frac ab+\frac cd=\frac{a\frac d{\gcd(b,d)}+c\frac b{\gcd(b,d)}}{\operatorname{lcm}(b,d)}$$and, since $a$, $\frac d{\gcd(b,d)}$, $c$, and $\frac b{\gcd(b,d)}$ are all integers, you can just take$$e=a\frac d{\gcd(b,d)}+c\frac b{\gcd(b,d)}\in\mathbb Z.$$