Consider the product of the following two partial series: $$\left[\sum_{n=1}^{N} \frac{1}{n^s}\right] \cdot \left[\sum_{d=1}^{N} \frac{1}{d^{s^*}}\right]$$ When $\mathbf{Re}(s) > 1$, then in the limit as $N \to \infty$, the above expression converges to $$\zeta(s) \cdot \zeta(s^*) = \zeta(s) \cdot \zeta(s)^* = |\zeta(s)|^2$$ Going back to the original partial series, we can rewrite the expression above as $$\sum_{(n,d) \in M_N} \left[\frac{1}{n^s} \cdot \frac{1}{d^{s^*}}\right]$$ where $M_N = \{(n,d) \in \Bbb N_+^2: \max(n,d) \leq N\}$, and $\Bbb N_+$ is the set of strictly positive naturals.
Again, if $\mathbf{Re}(s) > 1$, then in the limit as $N \to \infty$, the above series converges to $|\zeta(s)|^2$, since the series converges absolutely as explained in this answer.
My question is about changing the way the summation is bounded. Suppose that rather than bounding by $\max(n,d) \leq N$, we bound by $n \cdot d \leq N$. We can do so by defining the set $P_N = \{(n,d) \in \Bbb N_+^2: n \cdot d \leq N\}$, and looking at the series $$\sum_{(n,d) \in P_N} \left[\frac{1}{n^s} \cdot \frac{1}{d^{s^*}}\right]$$
Question: for $\mathbf{Re}(s) > 1$, does the above still converge to $|\zeta(s)|^2$ as $N \to \infty$?
It's kind of strange, because in the first case we could decompose the series bounded by $\max(n,d)\leq N$ into a product of two series from 1 to $N$, and then see that each factor converges to zeta as $N \to \infty$. If we're bounding by $n \cdot d$, we can't split the product in this way. However, the two series both converge to $$\sum_{(n,d) \in \Bbb N_+^2} \left[\frac{1}{n^s} \cdot \frac{1}{d^{s^*}}\right]$$ and since everything in sight is absolutely convergent, I would expect the two are equal and both converge to $|\zeta(s)|^2$.