While tinkering with the log-derivative of the function \begin{align}\varphi(z,q)&=2e^{-\pi qz^2-\pi q/4}\sinh\pi qz \prod_{k\ge1}(1-e^{-2k\pi q})(1-e^{-2k\pi q+2\pi qz})(1-e^{-2k\pi q-2\pi qz})\tag1\\ &=\frac2{\sqrt{q}}e^{-\tfrac{\pi}{4q}}\sin \pi z \prod_{k\ge1}(1-e^{-2k\pi/q})(1-e^{-2k\pi/q+2i\pi z})(1-e^{-2k\pi/q-2i\pi z})\tag2 \end{align} I found I could create a function around its moment integral (with an additional constant $\alpha$): $$u_n(z,\alpha;q)=\exp{\int_0^\alpha y^n \dfrac{\varphi'}{\varphi}(y+z,q)\,dy}\tag3$$ I also know that $\varphi(z+i/q,q)=-e^{\pi/q-2i\pi z}\varphi(z,q)$ and $\varphi(z+1,q)=-\varphi(z,q)$ (each following respectively from $(1)$ and $(2)$) and so $$\frac{\varphi'}{\varphi}(z+i/q,q)=-2i\pi+\frac{\varphi'}{\varphi}(z,q),\qquad\frac{\varphi'}{\varphi}(z+1,q)=\frac{\varphi'}{\varphi}(z,q)$$ If I plug these into $(3)$ I'll get \begin{align}u_n(z+i/q,\alpha,q)&=\exp\int_0^\alpha y^n\frac{\varphi'}{\varphi}(y+z+i/q,q)\,dy\\ &=\exp\int_0^\alpha y^n\left(-2i\pi+\frac{\varphi'}{\varphi}(y+z,q)\right)\,dy\\ &=e^{-2i\pi\alpha^{n+1}/(n+1)}u_n(z,\alpha,q)\end{align} and \begin{align}u_n(z+1,\alpha,q)&=\exp\int_0^\alpha y^n\frac{\varphi'}{\varphi}(y+z+1,q)\,dy\\ &=\exp\int_0^\alpha y^n\frac{\varphi'}{\varphi}(y+z,q)\,dy\\ &=u_n(z,\alpha,q)\end{align} But I've noticed that $\frac{\varphi(z+\alpha,q)}{\varphi(z,q)}$ satisfies similar properties: \begin{align}\frac{\varphi(z+\alpha+i/q,q)}{\varphi(z+i/q,q)}&=\frac{-e^{\pi/q-2i\pi(z+\alpha)}\varphi(z+\alpha,q)}{-e^{\pi/q-2i\pi z}\varphi(z,q)}=e^{-2i\alpha\pi}\frac{\varphi(z+\alpha,q)}{\varphi(z,q)}\\ \frac{\varphi(z+\alpha+1,q)}{\varphi(z+1,q)}&=\frac{-\varphi(z+\alpha,q)}{-\varphi(z,q)}=\frac{\varphi(z+\alpha,q)}{\varphi(z,q)}\end{align} So with the right exponents, I can combine them into a doubly periodic function:\begin{align}e_n(z,\alpha,q):&=\left[\frac{\varphi(z+\alpha,q)}{\varphi(z,q)}\right]^{\alpha^n}u_n^{-n-1}(z,\alpha,q)\tag4\\ &\\ e_n(z+1,\alpha,q)&=\left[\frac{-\varphi(z+\alpha,q)}{-\varphi(z,q)}\right]^{\alpha^n}u_n^{-n-1}(z+1,\alpha,q)\\ &=\left[\frac{\varphi(z+\alpha,q)}{\varphi(z,q)}\right]^{\alpha^n}u_n^{-n-1}(z,\alpha,q)\\ &=e_n(z,\alpha,q)\\ &\\ e_n(z+i/q,\alpha,q)&=\left[e^{-2\alpha i\pi}\frac{\varphi(z+\alpha,q)}{\varphi(z,q)}\right]^{\alpha^n}u_n^{-n-1}(z+i/q,\alpha,q)\\ &=e^{-2i\pi\alpha^{n+1}}\left[\frac{\varphi(z+\alpha,q)}{\varphi(z,q)}\right]^{\alpha^n}e^{2i\pi\alpha^{n+1}}u_n^{-n-1}(z,\alpha,q)\\ &=e_n(z,\alpha,q)\end{align} But the strangest property of $e_n$ follows from a transformation for $\varphi(z,q)$: $$\varphi(z,q)=-iq^{-1/2}e^{-\pi qz^2}\varphi(iqz,1/q)$$ which leads to \begin{align}e_n(z,\alpha,q)&=\left[\frac{-iq^{-1/2}e^{-\pi q(z+\alpha)^2}\varphi(iq(z+\alpha),1/q)}{-iq^{-1/2}e^{-\pi qz^2}\varphi(iqz,1/q)}\right]^{\alpha^n}\\ &\quad\exp\Big[-(n+1)\int_0^\alpha y^n\left\{-2\pi q(y+z)+iq\frac{\varphi'}{\varphi}(iqy+iqz,1/q)\right\}\,dy\Big]\\[2ex] &=e^{-\pi q\alpha^n(\alpha^2+2\alpha z)}\left[\frac{\varphi(iqz+\alpha iq,1/q)}{\varphi(iqz,1/q)}\right]^{\alpha^n}\\ &\quad\exp\Big[\left\{2\pi q\alpha^{n+2}\cdot\frac{n+1}{n+2}+2\pi qz\alpha^{n+1}-iq(n+1)\int_0^\alpha y^n\frac{\varphi'}{\varphi}(iqy+iqz,1/q)\,dy\right\}\Big]\\[2ex] &=e^{n\alpha^{n+1}\pi q/(n+2)}\left[\frac{\varphi(iqz+\alpha iq,1/q)}{\varphi(iqz,1/q)}\right]^{\alpha^n}\exp\int_0^{\alpha iq}-(n+1)(iq)^{-n}x^n\frac{\varphi'}{\varphi}(x+iqz,1/q),dx\\ &=e^{n\alpha^{n+1}\pi q/(n+2)}\left[\frac{\varphi(iqz+\alpha iq,1/q)}{\varphi(iqz,1/q)}\right]^{\alpha^n}u_n^{-(n+1)(iq)^{-n}}(iqz,\alpha iq,1/q)\\ &=e^{n\alpha^{n+1}\pi q/(n+2)}\left[\frac{\varphi(iqz+\alpha iq,1/q)}{\varphi(iqz,1/q)}\right]^{\alpha^n}\left[\frac{\varphi(iqz+\alpha iq, 1/q)}{\varphi(iqz,1/q)}\right]^{-\alpha^n}e_n^{(iq)^{-n}}(iqz,\alpha iq,1/q)\\ &=e^{\alpha^{n+1}n\pi q/(n+2)}e_n^{(iq)^{-n}}(iqz,\alpha iq,1/q) \end{align} I would like to know the degrees of the zeros and poles of $e_n$ so I can determine its Mittag-Leffler expansion. The above transformation appears to change their nature, so this may prove to be very thorny.
[Edit] I have found a tractable form for $e_1(z,\alpha;q)$ using the formula \begin{align}\exp\int_0^z\ln\varphi(y,q)\,dy=&e^{\pi qz^3/3-\pi z(q+q^{-1})/6}\varphi^z(z,q)\\ &\cdot\prod_{k\ge1}\underbrace{\bigg(\frac{1-e^{-2k\pi q-2\pi qz}}{1-e^{-2k\pi q+2\pi qz}}\bigg)^k}_{p_1(z,q)}\underbrace{\bigg(\frac{1-e^{-2k\pi/q+2i\pi z}}{1-e^{-2k\pi/q-2i\pi z}}\bigg)^{ik/q}}_{p_2(z,q)}:\tag5 \end{align} I've posted the details as an answer.
Using the formula \begin{align}\exp\int_0^z\ln\varphi(y,q)\,dy=&e^{\pi qz^3/3-\pi z(q+q^{-1})/6}\varphi^z(z,q)\\ &\cdot\prod_{k\ge1}\underbrace{\bigg(\frac{1-e^{-2k\pi q-2\pi qz}}{1-e^{-2k\pi q+2\pi qz}}\bigg)^k}_{p_1(z,q)}\underbrace{\bigg(\frac{1-e^{-2k\pi/q+2i\pi z}}{1-e^{-2k\pi/q-2i\pi z}}\bigg)^{ik/q}}_{p_2(z,q)}:\tag5 \end{align} \begin{align}e_1(z,\alpha;q)&=\left[\frac{\varphi(z+\alpha,q)}{\varphi(z,q)}\right]^\alpha\exp\Bigg[-2\int_0^\alpha y\frac{\varphi'}\varphi(y+z,q)\,dy\Bigg]\\[2ex] &=\left[\frac{\varphi(z+\alpha,q)}{\varphi(z,q)}\right]^\alpha\exp\Bigg[-2\bigg(\alpha\ln\varphi(z+\alpha,q)-\int_0^\alpha \ln\varphi(y+z,q)\,dy\bigg)\Bigg]\\[2ex] &=\varphi^{-\alpha}(z,q)\varphi^{-\alpha}(z+\alpha,q)\exp\bigg[2\int_z^{z+\alpha}\ln\varphi(u,q)\,du\bigg]\\ &=\varphi^{-\alpha}(z,q)\varphi^{-\alpha}(z+\alpha,q)\exp\bigg[\frac{2\alpha^3\pi q}3+2\alpha\pi qz(\alpha+z)-\frac{\alpha\pi}3\left(q+\frac1q\right)\bigg]\\ &\quad\cdot\frac{\varphi^{2z+2\alpha}(z+\alpha,q)}{\varphi^{2z}(z,q)}\prod_{k\ge1}\bigg(\frac{p_1(z+\alpha,q)p_2(z+\alpha,q)}{p_1(z,q)p_2(z,q)}\bigg)^2\\ &=\frac{\varphi^{2z+\alpha}(z+\alpha,q)}{\varphi^{2z+\alpha}(z,q)}e^{2\alpha^3\pi q/3+2\alpha\pi qz(\alpha+z)-\alpha\pi(q+q^{-1})/3}\prod_{k\ge1}\bigg(\frac{p_1(z+\alpha,q)p_2(z+\alpha,q)}{p_1(z,q)p_2(z,q)}\bigg)^2\end{align} To see if it holds, I'll need to calculate the $1$- and $i/q$-translates of $\prod_{k\ge1}p_1(z,q)p_2(z,q)$:\begin{align}\prod_{k\ge1}p_1(z+1,q)p_2(z+1,q)&=\prod_{k\ge1}\bigg(\frac{1-e^{-2k\pi q-2\pi q(z+1)}}{1-e^{-2k\pi q+2\pi q(z+1)}}\bigg)^k\bigg(\frac{1-e^{-2k\pi/q+2i\pi(z+1)}}{1-e^{-2k\pi/q-2i\pi(z+1)}}\bigg)^{ik/q}\\ &=\prod_{k\ge1}\bigg(\frac{1-e^{-(2k+2)\pi q-2\pi qz}}{1-e^{-(2k-2)\pi q+2\pi qz}}\bigg)^k\bigg(\frac{1-e^{-2k\pi/q+2i\pi z}}{1-e^{-2k\pi/q-2i\pi z}}\bigg)^{ik/q}\\[2ex] &=\prod_{k\ge1}p_2(z,q)\frac{\prod_{n\ge2}(1-e^{-2n\pi q-2\pi qz})^{n-1}}{\prod_{n\ge0}(1-e^{-2n\pi q+2\pi qz})^{n+1}}\\[2ex] &=\frac1{1-e^{2\pi qz}}\prod_{k\ge1}p_2(z,q)\prod_{n\ge1}\frac{p_1(z,q)}{(1-e^{-2n\pi q+2\pi qz})(1-e^{-2n\pi q-2\pi qz})}\\[2ex] &=\frac{2e^{-\pi qz^2-\pi q/4}\sinh\pi qz}{(1-e^{2\pi qz})\varphi(z,q)}\prod_{k\ge1}(1-e^{-2k\pi q})p_1(z,q)p_2(z.q)\\[2ex] &=-\frac{e^{-\pi qz^2-\pi qz-\pi q/4}}{\varphi(z,q)}\prod_{k\ge1}(1-e^{-2k\pi q})p_1(z,q)p_2(z.q) \end{align} Similarly,\begin{align}\prod_{k\ge1}p_1(z+i/q,q)p_2(z+i/q,q)&=\prod_{k\ge1}\bigg(\frac{1-e^{-2k\pi q-2\pi qz+2i\pi}}{1-e^{-2k\pi q+2\pi qz-2i\pi}}\bigg)^k\bigg(\frac{1-e^{-2k\pi/q+2i\pi(z+i/q)}}{1-e^{-2k\pi/q-2i\pi(z+i/q)}}\bigg)^{ik/q}\\ &=\prod_{k\ge1}p_1(z,q)\bigg(\frac{1-e^{-(2k+2)\pi/q+2i\pi z}}{1-e^{-(2k-2)\pi/q-2i\pi z}}\bigg)^{ik/q}\\[2ex] &=\prod_{k\ge1}p_1(z,q)\frac{\prod_{n\ge2}(1-e^{-2n\pi/q+2i\pi z})^{i/q(n-1)}}{\prod_{n\ge0}(1-e^{-2n\pi/q-2i\pi z})^{i/q(n+1)}}\\[2ex] &=\frac1{(1-e^{-2i\pi z})^{i/q}}\bigg(\frac{e^{-\pi/(4q)+i\pi z}(1-e^{-2i\pi z})}{i\sqrt q\varphi(z,q)}\prod_{k\ge1}(1-e^{-2k\pi/q})\bigg)^{i/q}\prod p_1(z,q)p_2(z,q)\\ &=\frac{e^{-i\pi/(4q^2)-\pi z/q}}{[i\sqrt q\varphi(z,q)]^{i/q}}\prod_{k\ge1}(1-e^{-2k\pi/q})^{i/q}p_1(z,q)p_2(z,q) \end{align} Now\begin{align}e_1(z+1,\alpha;q)&=\frac{\varphi^{2z+2+\alpha}(z+1+\alpha,q)}{\varphi^{2z+2+\alpha}(z+1,q)}\\ &\quad\cdot e^{2\alpha^3\pi q/3+2\alpha\pi q(z+1)(\alpha+z+1)-\alpha\pi(q+q^{-1})/3}\prod\bigg(\frac{p_1(z+\alpha+1,q)p_2(z+\alpha+1,q)}{p_1(z+1,q)p_2(z+1,q)}\bigg)^2\\ &=\frac{(-1)^{2z+2+\alpha}\varphi^{2z+2+\alpha}(z+\alpha,q)}{(-1)^{2z+\alpha+2}\varphi^{2z+2+\alpha}(z,q)}e^{2\alpha^3\pi q/3+2\alpha\pi qz(\alpha+z)-\alpha\pi(q+q^{-1})/3+2\alpha\pi q(2z+\alpha)+2\alpha\pi q}\\ &\quad\cdot\Bigg[\frac{-\varphi(z,q)e^{\pi qz^2+\pi qz+\pi q/4}}{-\varphi(z+\alpha,q)e^{\pi q(z+\alpha)^2+\pi q(z+\alpha)+\pi q/4}}\Bigg]^2\prod\bigg(\frac{p_1(z+\alpha,q)p_2(z+\alpha,q)}{p_1(z,q)p_2(z,q)}\bigg)^2\\[2ex] &=\bigg(\frac{\varphi(z+\alpha,q)}{\varphi(z,q)}\bigg)^2\bigg(\frac{-\varphi(z,q)}{-\varphi(z+\alpha,q)}\bigg)^2e^{4\alpha\pi qz+2\alpha^2\pi q+2\alpha\pi q-2(2\alpha\pi qz+\alpha^2\pi q+\alpha\pi q)}e_1(z,\alpha;q)\\[2ex] &=e_1(z,\alpha;q);\tag6 \end{align} \begin{align}e_1(z+i/q,\alpha;q)&=\frac{\varphi^{2z+2i/q+\alpha}(z+i/q+\alpha,q)}{\varphi^{2z+2i/q+\alpha}(z+i/q,q)}\\ &\quad\cdot e^{2\alpha^3\pi q/3+2\alpha\pi q(z+i/q)(\alpha+z+i/q)-\alpha\pi(q+q^{-1})/3}\prod\bigg(\frac{p_1(z+\alpha+i/q,q)p_2(z+\alpha+i/q,q)}{p_1(z+i/q,q)p_2(z+i/q,q)}\bigg)^2\\[2ex] &=\frac{[-e^{\pi/q-2i\pi(z+\alpha)}\varphi(z+\alpha,q)]^{2z+2i/q+\alpha}}{[-e^{\pi/q-2i\pi z}\varphi(z,q)]^{2z+\alpha+2i/q}}e^{2\alpha^3\pi q/3+2\alpha\pi qz(\alpha+z)-\alpha\pi(q+q^{-1})/3+2\alpha i\pi(\alpha+2z)-2\alpha\pi/q}\\ &\quad\cdot\bigg(\frac{i\sqrt q\varphi(z,q)}{i\sqrt q\varphi(z+\alpha,q)}\bigg)^{2i/q}e^{i\pi/(2q^2)+2\pi z/q-i\pi/(2q^2)-2\pi(z+\alpha)/q}\prod\bigg(\frac{p_1(z+\alpha,q)p_2(z+\alpha,q)}{p_1(z,q)p_2(z,q)}\bigg)^2\\[2ex] &=e^{-2\alpha i\pi(2z+2i/q+\alpha)+2\alpha i\pi(\alpha+2z)-4\alpha\pi/q}\bigg(\frac{\varphi(z+\alpha,q)}{\varphi(z,q)}\bigg)^{2i/q}\bigg(\frac{\varphi(z,q)}{\varphi(z+\alpha,q)}\bigg)^{2i/q}e_1(z,\alpha;q)\\[2ex] &=e_1(z,\alpha;q)\tag7 \end{align} $e_1$ is indeed elliptic! All that's left to know is if it has any zeros or singularities (though because I derived $(5)$ by expanding $\varphi^z(z,q)$ into a Weierstrass product, I doubt it.) Meanwhile, I'll try to extend this approach to higher $e_n$ with product expansions of $\varphi(z,q)^{z^n}$.