Theorem 329 of Hardy and Wright, An Introduction to the Theory of Numbers, says there is a positive constant $A$ such that $$A\lt{\sigma(n)\phi(n)\over n^2}\lt1$$
In a footnote, they show that $A=6/\pi^2$.
I don't know if we can use convolution to prove your statement, and it's the first time i see it so I don't know any reference , But I have a proof by induction.
The first terms $a_1,a_2$ coincide with Fibonacci numbers.
Let us suppose that:$a_1,a_2,\cdots,a_{x},a_{x+1}$ are the first Fibonacci numbers (this assumption is used in the proposition) and prove that $a_{x+2}$ is the next Fibonacci number.
we compute $a_{x+2}-a_{x+1}$:
$$a_{x+2}-a_{x+1}=\sum_{n:\alpha(n)=x} \varphi(n)+\sum_{n:\alpha(n)<x-1}\varphi(n) \left (\left \lfloor{ \frac{x}{\alpha(n)} }\right \rfloor - \left \lfloor{ \frac{x-1}{\alpha(n)} }\right \rfloor\right )$$
And we know that $ \lfloor x/k \rfloor - \lfloor (x-1)/k \rfloor = 1 $ when $k|x$ (for $k\geq1$) and $0$ otherwise, so
$$a_{x+2}-a_{x+1}=\sum_{n:\alpha(n)|x} \varphi(n)\,\,\,\, (1)$$
but we have
Proposition for every positive integer $n$ $$\alpha(n)|x \Leftrightarrow n|a_x$$
proof
first it's clear that if $\alpha(n)=k$ with $k|x$ then $n|a_k$, and because $a_k$ and $a_x$ are Fibonacci numbers then $a_k|a_x$ finally $n|a_x$.
Second, given a divisor $n$ of $a_x$,let $\alpha(n)=k\leq x$ then $n|a_k$ so $n|gcd(a_k,a_x)=a_{gcd(x,k)}$ ( because $a_x$ and $a_k$ are Fibonacci numbers) using the definition of $\alpha(n)$ we have $k\leq gcd(x,k)$ hence $k|x$.
Using the proposition, the sum $(1)$ becomes:
$$a_{x+2}-a_{x+1}=\sum_{n:n|a_x} \varphi(n)$$
using Euler's identity $\sum_{d:d|n} \varphi(d)=n$ :
$$a_{x+2}-a_{x+1}=a_x $$
Finally $a_{x+2}$ is the next Fibonacci number
Best Answer
$(\ldots \text{continued from comments})$ Consider an example when $\color{blue}{n=10}$.
Divisors of $\color{blue}{10}$ are $\{1,2,5,10\}$
$\varphi(1)=1$
$\varphi(2)=1$
$\varphi(5)=4$
$\varphi(10)=4$
Therefore $$\begin{align}\sum_{k|10} \varphi(k) &= \varphi(1)+ \varphi(2)+ \varphi(5)+ \varphi(10) \\~\\&=1+1+4+4\\~\\&=\color{blue}{10}\end{align}$$
as desired.