What's the sum of the reciprocals of the numbers that can be written as the sum of two positive cubes?

A very specific question: What's the sum of the reciprocals of the numbers that can be written as the sum of two positive cubes (Oeis: A003325)?

$$\begin{align} &\sum_{n=1}^\infty \frac{1}{A003325(n)} \\ & =\frac{1}{1^3+1^2}+\frac{1}{1^3+2^3}+\frac{1}{2^3+2^3}+\frac{1}{1^3+3^3}+\frac{1}{2^3+3^3}+\frac{1}{3^3+3^3}+\frac{1}{1^3+4^3}+\dots \\ & = \frac{1}{2}+\frac{1}{9}+\frac{1}{16}+\frac{1}{28}+\frac{1}{35}+\frac{1}{54}+\frac{1}{65}+\frac{1}{72}+\frac{1}{91}+\dots \end{align} $$

Some notation to organize my thinking:

Let $S\subset \mathbb{N}$ and let $1_{S}(x)=\cases{1 \mbox{ when }{x\in S }\\0 \mbox{ when } x\notin S}$

And define $m(S)=\sum_{n=1}^\infty \frac{1_S(n)}{n}$. This would be the sum of the reciprocals of $S$.

$S_{a,b}=\{n\in \mathbb{N}: \exists \vec{x} \in\mathbb{N^{a}}, \sum_{i=1}^a |x_i|^b=n \}$. This is the set of numbers that can be written as sum of $a$ positive numbers raised to the fixed power $b$.

Some immediate results of this notation. And some examples.

$m(S_{1,k})=\zeta(k)$ and $m(S_{4,2})$ is a divergent sum because $S_{4,2}=\mathbb{N}$.

My question above is asking for $m(S_{2,3})$. And somewhat more broadly I am asking: Can we get a handle on the density of this set? Why should we have any chances of answering this?

As we can read here, here, here and here. The first two links are MSE questions and the latter two are publications by Kevin A. Broughan.

$$n\in S_{2,3} \iff \exists m \mid n ,\quad n^{1/3} \leq m \leq 4^{1/3} n^{1/3} \mbox{ s.t. }\\ ( m^{2} - \frac{n}{m})=3l \mbox{ and }(m^{2} - 4l) \mbox{ is a perfect square. }$$

So because we have this nice characterization I was curious if this number could possibly be expressed in relationship to other well-known constants like value of the zeta function.

Note that the related question: What's the sum of the reciprocals of the numbers that can be written as the sum of two non-negative cubes? But we can see that this question is just $\zeta(3)$ away from the title question. That is,
$$m(\{n:(x,y)\in \mathbb{N_0^2}: n=x^3+y^3 \})=m(S_{2,3})+\zeta(3)$$

Here is what I know so far $m(S_{2,3}) \approx 0.9777693455 =\sum_{n=1}^{20000} \frac{1}{A003325(n)}$.

Can anyone provide any more insight into what this number is?

Apologies

Note that in the original version of this I made a typo and wrote $S_{a,b}=\{n\in \mathbb{N}: \exists \vec{x} \in\mathbb{Z^{a}}, \sum_{i=1}^a |x_i|^b=n \}$ this isn't quite what I wanted to type here.

I would like to mention that from Parseval theorem and the reflection formula we can achieve (though perhaps not to our full liking) a formulation for $L$ defined below which is related to the OP's query but not quite the same. $$L=\sum_{n,m \ge 1} \frac1{n^3+m^3} = \int_0^\infty f(x)^2dx= \frac1{2\pi} \int_{-\infty}^\infty |F(1/2+it)|^2dt$$ $$=\frac12 \int_{-\infty}^\infty \frac{|\zeta(3/2+3it)|^2}{|\sin(\pi /2+i\pi t)|}dt=\frac12 \int_{-\infty}^\infty \frac{|\zeta(3/2+3it)|^2}{\cosh(\pi t)}dt$$

Unfortunately the residue theorem doesn't apply.

$$f(x)= \sum_{n\ge 1} e^{-n^3x},\qquad F(s)=\Gamma(s)\zeta(3s)= \int_0^\infty x^{s-1}f(x)dx$$

The $L$ defined above should be approximately $2m(S_{2,3})-\zeta(3)$