1. Let me address the case of $z=1+i$. It seems to me that those are the only counterexamples, but a proof I can imagine is now hidden under technical details.

Consider the representation $z=ab+cd$ and define integers $\alpha=|a|^2$, $\beta=|b|^2$, $\gamma=|c|^2$, $\delta=|d|^2$. WLOG we have $$ \alpha\geq \beta>\gamma\geq\delta. $$

Let the vectors $\overrightarrow{OA}$ and $\overrightarrow{OB}$ represent $z$ (fixed) and $ab$ (variable), so that the vector $\overrightarrow{BA}$ represents $cd$. The sides of $OAB$ are $\sqrt2$, $\sqrt{\alpha\beta}$ and $\sqrt{\gamma\delta}$.

Assume that $\alpha\geq \gamma+2$. Then we have $$ \sqrt2 \geq \sqrt{\alpha\beta}-\sqrt{\gamma\delta}\geq \sqrt{(\gamma+1)(\gamma+2)}-\gamma, $$ so $$ (\gamma+1)(\gamma+2)\leq (\gamma+\sqrt2)^2, $$ which provides an explicit upper bound on $\gamma$ (and then a bound on $\alpha$ is also easy).

A similar argument works if $\delta\leq \beta-2$.

It remains to check the case $\alpha=\beta=\gamma+1=\delta+1$. Let $BH$ be an altitude of the triangle. Then $H$ represents an integer multiple of $\frac{1+i}2$, so $BH$ is an integer multiple of $\frac1{\sqrt2}$. However, $$ 2AH\pm\sqrt2=OH+AH=\frac{OH^2-AH^2}{OH-AH}=\frac{OB^2-AB^2}{\sqrt2}=\pm\frac{2\gamma+1}{\sqrt2}, $$ so that $BH$ is an odd multiple of $\frac1{2\sqrt2}$. This contradiction finishes the proof.

2. Some remarks on what to do in the general case.

The first part can be applied similarly, yielding that, if $S(z)$ is infinite, then there are infinitely many solutions with small and fixed values of $\alpha-\beta$, $\beta-\gamma$, and $\gamma-\delta$.

But it seems that one such series should exist. This boils down to some equations similar in vein to Pell equations, which I have no time to dig through right now, sorry.

1. Let me address the case of $z=1+i$. It seems to me that those are the only counterexamples, but a proof I can imagine is now hidden under technical details.

Consider the representation $z=ab+cd$ and define integers $\alpha=|a|^2$, $\beta=|b|^2$, $\gamma=|c|^2$, $\delta=|d|^2$. WLOG we have $$ \alpha\geq \beta>\gamma\geq\delta. $$

Let the vectors $\overrightarrow{OA}$ and $\overrightarrow{OB}$ represent $z$ (fixed) and $ab$ (variable), so that the vector $\overrightarrow{BA}$ represents $cd$. The sides of $OAB$ are $\sqrt2$, $\sqrt{\alpha\beta}$ and $\sqrt{\gamma\delta}$.

Assume that $\alpha\geq \gamma+2$. Then we have $$ \sqrt2 \geq \sqrt{\alpha\beta}-\sqrt{\gamma\delta}\geq \sqrt{(\gamma+1)(\gamma+2)}-\gamma, $$ so $$ (\gamma+1)(\gamma+2)\leq (\gamma+\sqrt2)^2, $$ which provides an explicit upper bound on $\gamma$ (and then a bound on $\alpha$ is also easy).

A similar argument works if $\delta\leq \beta-2$.

It remains to check the case $\alpha=\beta=\gamma+1=\delta+1$. Let $BH$ be an altitude of the triangle. Then $H$ represents an integer multiple of $\frac{1+i}2$, so $AH$ is an integer multiple of $\frac1{\sqrt2}$. However, $$ 2AH\pm\sqrt2=OH+AH=\frac{OH^2-AH^2}{OH-AH}=\frac{OB^2-AB^2}{\sqrt2}=\pm\frac{2\gamma+1}{\sqrt2}, $$ so that $AH$ is an odd multiple of $\frac1{2\sqrt2}$. This contradiction finishes the proof.

2. Some remarks on what to do in the general case.

The first part can be applied similarly, yielding that, if $S(z)$ is infinite, then there are infinitely many solutions with small and fixed values of $\alpha-\beta$, $\beta-\gamma$, and $\gamma-\delta$.

But it seems that one such series should exist. This boils down to some equations similar in vein to Pell equations, which I have no time to dig through right now, sorry.