In this answer, we give a heuristic explanation for the formula (18) in Ref. 1.

1. Consider two zeta-function regularized infinite products $$\tag{1a} F_c(b)~:=~\prod_{\lambda \in \mathbb{N}+b} c~=~c^{-\frac{1}{2}-b} $$ and $$\tag{1b} G(b)~:=~\prod_{\lambda \in \mathbb{N}+b}\lambda~=~\frac{\sqrt{2\pi}}{\Gamma(b+1)} $$ over a half-lattice $\Lambda= \mathbb{N}+b$, where $b,c\in\mathbb{C}$.

2. Notice that if we put $b=0$, we get the well-known zeta function regularization formulas $$\tag{2a} F_c(b\!=\!0)~:=~\prod_{n \in \mathbb{N}} c~=~c^{-\frac{1}{2}} $$ and $$\tag{2b} G(b\!=\!0)~:=~\prod_{n \in \mathbb{N}}n~=~\sqrt{2\pi}. $$

3. Using OP's argument, one would falsely have expected that the infinite product (1a) should be independent of the shift $b\in\mathbb{C}$. Here we will employ another kind of logic. If we shift the half-lattice $\Lambda= \mathbb{N}+b$ by one unit $b\to b-1$, we would expect one more element in the half-lattice, and thereby one more factor in the infinite product. This leads to the following very powerful functional equations/recursion relations, $$\tag{3a} F_c(b-1)~=~c F_c(b) $$ and $$\tag{3b} G(b-1)~=~b G(b). $$ Eqs. (3a) and (3b) are in fact satisfied by the zeta function regularization (1a) and (1b), respectively!

4. Therefore we naturally arrive at the regularized product formula (18) in Ref. 1, $$\prod_{n \in \mathbb{N}}(un+a) ~=~\prod_{\lambda \in \mathbb{N}+\frac{a}{u}} u\lambda ~=~\left[\prod_{\lambda \in \mathbb{N}+\frac{a}{u}} u\right]\left[\prod_{\lambda \in \mathbb{N}+\frac{a}{u}}\lambda\right]$$ $$\tag{18}~=~F_u\left(\frac{a}{u}\right)G\left(\frac{a}{u}\right) ~=~u^{-\frac{1}{2}-\frac{a}{u}}\frac{\sqrt{2\pi}}{\Gamma\left(\frac{a}{u}+1\right)}.$$

References:

1. J.R. Quine, S.H. Heydari and R.Y. Song, Zeta regularized products, Trans. Amer. Math. Soc. 338 (1993) 213; eq. (18).

In this answer, we give a heuristic explanation for the formula (18) in Ref. 1.

1. Consider two zeta-function regularized infinite products $$\tag{1a} F_a(b)~:=~\prod_{\lambda \in \mathbb{N}+b} a~=~a^{-\frac{1}{2}-b} $$ and $$\tag{1b} G(b)~:=~\prod_{\lambda \in \mathbb{N}+b}\lambda~=~\frac{\sqrt{2\pi}}{\Gamma(b+1)} $$ over a half-lattice $\Lambda= \mathbb{N}+b$. Here $a\in\mathbb{C}\backslash\{0\}$ and $b\in\mathbb{C}$ are two complex numbers.

2. Notice that if we put $b=0$, we get the well-known zeta function regularization formulas $$\tag{2a} F_a(b\!=\!0)~:=~\prod_{n \in \mathbb{N}} a~=~a^{-\frac{1}{2}} $$ and $$\tag{2b} G(b\!=\!0)~:=~\prod_{n \in \mathbb{N}}n~=~\sqrt{2\pi}. $$

3. Using OP's argument, one would falsely have expected that the infinite product (1a) should be independent of the shift $b\in\mathbb{C}$. Here we will employ another kind of logic. If we shift the half-lattice $\Lambda= \mathbb{N}+b$ by one unit $b\to b-1$, we would expect one more element in the half-lattice, and thereby one more factor in the infinite product. This leads to the following very powerful functional equations/recursion relations, $$\tag{3a} F_a(b\!-\!1)~=~a ~ F_a(b) $$ and $$\tag{3b} G(b\!-\!1)~=~b ~ G(b). $$ Note in particular that eqs. (3a) and (3b) are in fact satisfied by the zeta function regularization (1a) and (1b), respectively!

4. Therefore we naturally arrive at the regularized product formula (18) in Ref. 1, $$\prod_{n \in \mathbb{N}}(an+b) ~=~\prod_{\lambda \in \mathbb{N}+\frac{b}{a}} a\lambda ~=~\left[\prod_{\lambda \in \mathbb{N}+\frac{b}{a}} a\right]\left[\prod_{\lambda \in \mathbb{N}+\frac{b}{a}}\lambda\right]$$ $$\tag{18}~=~F_a\left(\frac{b}{a}\right)G\left(\frac{b}{a}\right) ~=~a^{-\frac{1}{2}-\frac{b}{a}}\frac{\sqrt{2\pi}}{\Gamma\left(\frac{b}{a}+1\right)}.$$

References:

1. J.R. Quine, S.H. Heydari and R.Y. Song, Zeta regularized products, Trans. Amer. Math. Soc. 338 (1993) 213; eq. (18).

In this answer, we give a heuristic explanation for the formula in Ref. 1.

1. Consider two zeta-function regularized infinite products $$\tag{1a} F_c(b)~:=~\prod_{\lambda \in \mathbb{N}+b} c~=~c^{-\frac{1}{2}-b} $$ and $$\tag{1b} G(b)~:=~\prod_{\lambda \in \mathbb{N}+b}\lambda~=~\frac{\sqrt{2\pi}}{\Gamma(b+1)} $$ over a half-lattice $\Lambda= \mathbb{N}+b$, where $b,c\in\mathbb{C}$.

2. Notice that if we put $b=0$, we get the well-known zeta function formulas $$\tag{2a} F_c(b\!=\!0)~:=~\prod_{n \in \mathbb{N}} c~=~c^{-\frac{1}{2}} $$ and $$\tag{2b} G(b\!=\!0)~:=~\prod_{n \in \mathbb{N}}n~=~\sqrt{2\pi}. $$

3. If we shift the half-lattice $\Lambda= \mathbb{N}+b$ by one unit $b\to b-1$, we would expect the following very powerful functional equations/recursion relations, $$\tag{3a} F_c(b-1)~=~c F_c(b) $$ and $$\tag{3b} G(b-1)~=~b G(b). $$ Eqs. (3a) and (3b) are in fact satisfied by the zeta function regularization (1a) and (1b), respectively!

4. Therefore we naturally arrive at the regularized product formula in Ref. 1, $$\prod_{n \in \mathbb{N}}(un+a) ~=~\prod_{\lambda \in \mathbb{N}+\frac{a}{u}} u\lambda ~=~\left[\prod_{\lambda \in \mathbb{N}+\frac{a}{u}} u\right]\left[\prod_{\lambda \in \mathbb{N}+\frac{a}{u}}\lambda\right]$$ $$\tag{4}~=~F_u\left(\frac{a}{u}\right)G\left(\frac{a}{u}\right) ~=~u^{-\frac{1}{2}-\frac{a}{u}}\frac{\sqrt{2\pi}}{\Gamma\left(\frac{a}{u}+1\right)}.$$

References:

1. J.R. Quine, S.H. Heydari and R.Y. Song, Zeta regularized products, Trans. Amer. Math. Soc. 338 (1993) 213; eq. (18).

In this answer, we give a heuristic explanation for the formula (18) in Ref. 1.

1. Consider two zeta-function regularized infinite products $$\tag{1a} F_c(b)~:=~\prod_{\lambda \in \mathbb{N}+b} c~=~c^{-\frac{1}{2}-b} $$ and $$\tag{1b} G(b)~:=~\prod_{\lambda \in \mathbb{N}+b}\lambda~=~\frac{\sqrt{2\pi}}{\Gamma(b+1)} $$ over a half-lattice $\Lambda= \mathbb{N}+b$, where $b,c\in\mathbb{C}$.

2. Notice that if we put $b=0$, we get the well-known zeta function regularization formulas $$\tag{2a} F_c(b\!=\!0)~:=~\prod_{n \in \mathbb{N}} c~=~c^{-\frac{1}{2}} $$ and $$\tag{2b} G(b\!=\!0)~:=~\prod_{n \in \mathbb{N}}n~=~\sqrt{2\pi}. $$

3. Using OP's argument, one would falsely have expected that the infinite product (1a) should be independent of the shift $b\in\mathbb{C}$. Here we will employ another kind of logic. If we shift the half-lattice $\Lambda= \mathbb{N}+b$ by one unit $b\to b-1$, we would expect one more element in the half-lattice, and thereby one more factor in the infinite product. This leads to the following very powerful functional equations/recursion relations, $$\tag{3a} F_c(b-1)~=~c F_c(b) $$ and $$\tag{3b} G(b-1)~=~b G(b). $$ Eqs. (3a) and (3b) are in fact satisfied by the zeta function regularization (1a) and (1b), respectively!

4. Therefore we naturally arrive at the regularized product formula (18) in Ref. 1, $$\prod_{n \in \mathbb{N}}(un+a) ~=~\prod_{\lambda \in \mathbb{N}+\frac{a}{u}} u\lambda ~=~\left[\prod_{\lambda \in \mathbb{N}+\frac{a}{u}} u\right]\left[\prod_{\lambda \in \mathbb{N}+\frac{a}{u}}\lambda\right]$$ $$\tag{18}~=~F_u\left(\frac{a}{u}\right)G\left(\frac{a}{u}\right) ~=~u^{-\frac{1}{2}-\frac{a}{u}}\frac{\sqrt{2\pi}}{\Gamma\left(\frac{a}{u}+1\right)}.$$

References:

1. J.R. Quine, S.H. Heydari and R.Y. Song, Zeta regularized products, Trans. Amer. Math. Soc. 338 (1993) 213; eq. (18).