For a multiplicative arithmetic function $f(n)$, we define $ F(s) = \sum_{n\ge1}^{} \dfrac{f(n)}{n^s}$. We then define the coefficients $\Lambda_f (n)$ by $$ -\dfrac{F'(s)}{F(s)} = \sum_{n\ge1}^{} \dfrac{\Lambda_f (n)}{n^s}$$ It's easy to see that the $\Lambda_f (n) $s are non-zero only at prime powers. We also have the formula $$ f(n)\log n =\sum_{d \lvert n}^{} \Lambda_f (d) f(n/d)$$
Now we need to show that if $\lvert f(n) \rvert \le 1, \forall n$, then there are constants $A, C$ such that $\sum_{m\le z}^{} \lvert \Lambda_f (n) \rvert \le Az + C$, for all $z \ge 1$. I can show this when $f$ is completely multiplicative (using an explicit formula for $\Lambda_f (n)$), but I'm stuck in the more general case, when $f$ is just multiplicative. Any suggestions?
If $|f(n)|\leq 1$ then $\sum_{n\geq 1}\frac{f(n)}{n^s}$ is holomorphic over $\text{Re}(s)>1$ and over such region we have $$ F(s) = \prod_{p}\left(1+\frac{f(p)}{p^s}+\frac{f(p^2)}{p^{2s}}+\ldots\right) \tag{1}$$ $$ -\frac{F'(s)}{F(s)}=\sum_{p}\frac{1+\frac{f(p)\log p}{p^s}+\frac{f(p^2)\log p^2}{p^{2s}}+\ldots}{1+\frac{f(p)}{p^s}+\frac{f(p^2)}{p^{2s}}+\ldots}=\sum_{n\geq 1}\frac{\Lambda_f(n)}{n^s}\tag{2}$$ and since $F(s)\cdot\frac{F'(s)}{F(s)}=F'(s) = -\sum_{n\geq 1}\frac{f(n)\log n}{n^s}$ we have $ f(n) \log(n) = (\Lambda_f*f)(n)$.
Notice that if $F(s)$ has a pole at $s=1$, then $-\frac{F'(s)}{F(s)}$ has a simple pole at $s=1$, and by applying Abel's summation to the RHS of $(2)$ one gets that the sums $\sum_{n\leq x}\left|\Lambda_f(n)\right|$ cannot grow too fast, namely not faster than a linear polynomial.