Can we invert these analogous "Dirichlet" series for GCD / LCM convolution?

This is probably not an answer to your current question, but for what it's worth:

I have written up a proof of the fact that the lcm-convolution of two multiplicative arithmetic functions is multiplicative (and I don't mean the proof I outlined in my comments; it's a different, cleaner proof). But I've also discovered that this is a result of von Sterneck and Lehmer from the $\leq$1930s. See Theorem 2.10.4 in my 18.781 (Spring 2016): Floor and arithmetic functions and references therein. The main vehicle of the proof is Theorem 2.10.5, which can be restated as follows:

Notations.

• Let $A$ be the $\mathbb{C}$-algebra of all arithmetic functions (i.e., functions from $\left\{1,2,3,\ldots\right\}$ to $\mathbb{C}$).

• Let $\star$ be the Dirichlet convolution on $A$; this is the binary operation on $A$ defined by $\left(f \star g\right)\left(n\right) = \sum\limits_{d \mid n} f\left(d\right) g\left(\dfrac{n}{d}\right) = \sum\limits_{\substack{d \geq 1;\ e \geq 1; \\ de = n}} f\left(d\right) g\left(e\right)$ for every two arithmetic functions $f$ and $g$ and every positive integer $n$.

• Let $\widetilde{\star}$ be the "lcm"-convolution on $A$; this is the binary operation on $A$ defined by $\left(f \widetilde{\star} g\right)\left(n\right) = \sum\limits_{\substack{d \geq 1;\ e \geq 1; \\ \operatorname{lcm}\left(d, e\right) = n}} f\left(d\right) g\left(e\right)$ for every two arithmetic functions $f$ and $g$ and every positive integer $n$.

• Let $\cdot$ be the pointwise product on $A$; this is the binary operation on $A$ defined by $\left(f \cdot g\right) \left(n\right) = f\left(n\right) g\left(n\right)$ for every two arithmetic functions $f$ and $g$ and every positive integer $n$.

Theorem A. Let $D$ be the map that sends each arithmetic function $f$ to the arithmetic function $F$ defined by $F\left(n\right) = \sum\limits_{d \mid n} f\left(d\right)$. (Note that $F$ can also be described as $\underline{1} \star f$, where $\underline{1}$ is the arithmetic function that is constantly $1$.) Then, $D$ is an isomorphism from the $\mathbb{C}$-algebra $\left(A, \widetilde{\star}\right)$ to the $\mathbb{C}$-algebra $\left(A, \cdot\right)$.

The map $F$ and its inverse both preserve multiplicativity (in fact, $F$ is Dirichlet convolution with the multiplicative function $\underline{1}$, whereas its inverse $F^{-1}$ is Dirichlet convolution with the multiplicative Möbius function $\mu$); thus, it is easy to see that the $\widetilde{\star}$ operation preserves multiplicativity. I don't say the word "isomorphism" in my note, since it is written for a pre-abstract-algebra audience, but what I do is a fairly transparent back-and-force argument using $F$ and $F^{-1}$.

Note that the $\mathbb{C}$-algebras $\left(A, \widetilde{\star}\right)$ and $\left(A, \cdot\right)$ are mutually isomorphic, but they are not isomorphic to the $\mathbb{C}$-algebra $\left(A, \star\right)$. Indeed, the former two algebras are isomorphic to the direct product $\prod_{n \geq 1} \mathbb{C}$ and thus are not integral domains, whereas the latter algebra is an integral domain (this can be proven by the same argument that one uses to show that formal power series over an integral domain form an integral domain: namely, if $f \in A$ and $g \in A$ are both nonzero, then we can pick a minimal $d \geq 1$ such that $f\left(d\right) \neq 0$ and a minimal $e \geq 1$ such that $g\left(e\right) \neq 0$; then, $\left(f \star g\right) \left(de\right) = f\left(d\right) g\left(e\right) \neq 0$ and thus $f \star g \neq 0$).

Now, you are trying to define a "gcd-convolution" on $A$, which should be a binary operation $\#$ satisfying $\left(f \# g\right)\left(n\right) = \sum\limits_{\substack{d \geq 1;\ e \geq 1; \\ \gcd\left(d, e\right) = n}} f\left(d\right) g\left(e\right)$ for every two arithmetic functions $f$ and $g$ and every positive integer $n$ for which the sum converges. As you have noticed, the sum does not always converge, and it is not clear what kind of convergence is the right kind to ask for. I wouldn't be surprised that if you allow conditional convergence, the $\#$ convolution won't even be associative.

The easiest way to avoid convergence questions is to restrict yourself to finitely supported arithmetic functions -- i.e., arithmetic functions $f$ for which the set $\left\{n \geq 1 \mid f\left(n\right) \neq 0\right\}$ is finite. It is easy to see that if $f$ and $g$ are two finitely supported arithmetic functions, then the arithmetic function $f \# g$ is well-defined and also finitely supported. Thus, if $A_0$ denotes the subspace of $A$ consisting of all finitely supported arithmetic functions, then $\left(A_0, \#\right)$ is a nonunital $\mathbb{C}$-algebra. Note that $\left(A_0, \star\right)$ and $\left(A_0, \cdot\right)$ and $\left(A_0, \widetilde{\star}\right)$ are nonunital $\mathbb{C}$-algebras as well. Moreover, the two nonunital $\mathbb{C}$-algebras $\left(A_0, \cdot\right)$ and $\left(A_0, \#\right)$ are isomorphic:

Theorem B. Let $U$ be the map that sends each arithmetic function $f \in A_0$ to the arithmetic function $F$ defined by $F\left(n\right) = \sum\limits_{n \mid d} f\left(d\right)$ (where the sum ranges over all positive multiples $d$ of $n$). Then, $U$ is an isomorphism from the nonunital $\mathbb{C}$-algebra $\left(A_0, \#\right)$ to the nonunital $\mathbb{C}$-algebra $\left(A_0, \cdot\right)$.

The two isomorphic nonunital rings $\left(A_0, \#\right)$ and $\left(A_0, \cdot\right)$ have no unity, while the two rings $\left(A_0, \widetilde{\star}\right)$ and $\left(A_0, \star\right)$ have a unity (namely, in both cases, the arithmetic function $\varepsilon$ that sends $1$ to $1$ and all larger integers to $0$). Thus, the two former rings are not isomorphic to any of the latter. Moreover, the two latter rings $\left(A_0, \widetilde{\star}\right)$ and $\left(A_0, \star\right)$ are not isomorphic either. Indeed, the ring $\left(A_0, \star\right)$ is an integral domain (being a subring of the integral domain $\left(A, \star\right)$), whereas the ring $\left(A_0, \widetilde{\star}\right)$ is not (for example: pick two distinct primes $p$ and $q$; now, for each positive integer $h$, let $e_h \in A_0$ be the arithmetic function sending $h$ to $1$ and all other positive integers to $0$; then $\left(e_p - e_{pq}\right) \widetilde{\star} e_q = 0$ but $e_p - e_{pq} \neq 0$ and $e_q \neq 0$).