Yes, this can be done. Let $K$ be a $S_4$-quartic field, $C$ its cubic resolvent field, and $L$ the Galois closure of $K$. By Galois correspondence, $L$ contains a unique quadratic subfield $Q$ which corresponds to the alternating group $A_4$; by our hypothesis, this quadratic subfield is equal to $\mathbb{Q}(\sqrt{-1})$. Note that the Galois closure $C^\prime$ of $C$ is also contained in $L$, and by Galois correspondence again $C^\prime$ has a unique quadratic subfield, which is then necessarily equal to $Q = \mathbb{Q}(\sqrt{-1})$. Therefore, to construct the required quartic polynomials (fields) we first look at the possible cubic resolvent fields.

It is known that the discriminant of a cubic field $C$ can be expressed uniquely in the form $\Delta(C) = df^2$, where $d$ is the discriminant of the quadratic resolvent field of $C$. By our assumption, we must have $d = -4$. We are thus looking for cubic fields whose discriminants are equal to $-(2k)^2$ for some $k \geq 1$. By the Delone-Faddeev correspondence, this is the same as looking for ($\text{GL}_2(\mathbb{Z})$-equivalence classes of) binary cubic forms with integer coefficients and discriminant $-4k^2$.

Restricting to monic binary cubic forms of the shape $F(x,y) = x^3 - Axy^2 + By^3$ whose discriminant is $4A^3 - 27B^2$, this gives the equation

$$\displaystyle 4A^3 = 27B^2 - 4k^2 \text{ giving } A^3 = 27b^2 - k^2, b = B/2.$$

This equation defines a genus 0 curve and can be explicitly parametrized.

Now to answer the question: given an initial $F_1$ with discriminant $-k^2$, construct infinitely many cubic fields $C_\ell$ with discriminant equal to $-4\ell^2$ and $\gcd(k, \ell)$ equal a power of 2 as above. Then, using the results of this paper (Dirichlet series associated to quartic fields with given cubic resolvent) by Cohen and Thorne, one can give infinitely many quartic fields with $K$ with cubic resolvent field equal to $C_\ell$. The resulting Galois closures of the fields $K$ generated this way will satisfy your constraints.