This question is an attempt to complete the issues discussed in a previous question of mine (How did Gauss sum Eisenstein series?), since my updated question did not recieve any attention. In my previous question, I mentioned that Gauss wrote an infinite series for the logarithm of the denominator of $\mathbb{sinlemn}(z) = \frac{M(z)}{N(z)}$. Since
$$N(z) = \prod (1-\frac{z}{((m+\frac{1}{2})+(n+\frac{1}{2})i)\varpi})$$
(that is, $\mathbb{sinlemn}(z)$ has poles at Gaussian half-integers multiples of $\varpi$). Gauss wrote:
$$\mathbb{log}N(z) =\frac{1}{12}z^4 - \frac{1}{280}z^8 +\frac{1}{4950}z^{12} - ...$$.
Since the logarithm of a infinite product equals an infinite sum of logarithms, one gets that:
$$\mathbb{log}N(z) = \sum \mathbb{log}(1-\frac{z}{((m+\frac{1}{2})+(n+\frac{1}{2})i)\varpi})$$,
and by the taylor series expansion of $\mathbb{log}(1-z)$ one gets that Gauss's infinite series for $\mathbb{log}N(z)$ is equivalent to the summation of a kind of "generalized Eisenstein series" in which the lattice is shifted by $(\frac{1}{2}+\frac{1}{2}i)\varpi$. Since such shifted lattice cannot be generated by some action of the modular group on the lattice of Gaussian integers (if it was, one could use the modularity of the Eisenstein series and deduce the series for $\mathbb{log}N(z)$ from that of $\mathbb{log}M(z)$), I wondered what tools enable to sum such series and what was Gauss's original method in this case. I tried to make a Google search about "modular forms defined on shifted lattices", but without success.
Side remark
I have no intention to "spam" StackExchange Mathematics with multitudes of more or less similar questions, so I have no problem to close this question and instead get an answer to my original (updated) question, if other users will vote to do so.
With $$M(z)=\prod_{a,b\in \Bbb{Z}^2-0} (1-\frac{z/\varpi}{a+ib}), \qquad N(z)=\prod_{a,b\in \Bbb{Z}^2} (1-\frac{z/\varpi}{a+ib+(1+i)/2})$$ (grouping the $a,b$ term with the $\pm a,\pm b$ terms to ensure the convergence)
Then $$N(z/2) = \prod_{a,b\in \Bbb{Z}^2-0,a=b=1\bmod 2} (1-\frac{z/\varpi}{a+ib})$$ $$M(z/2) = \prod_{a,b\in \Bbb{Z}^2-0,a=b=0\bmod 2} (1-\frac{z/\varpi}{a+ib})$$ $$ M(z/(1+i)) =\prod_{a,b\in \Bbb{Z}^2-0,a=b\bmod 2} (1-\frac{z/\varpi}{a+ib})$$ So that $$M(z/2)N(z/2)=M(z/(1+i))$$