Do $|s(t)|$ and $|S(f)|$ uniquely determine $s(t)$?
As pointed out in a comment by Jazzmaniac, the proof below is flawed. The mistake I made was the assumption leading to Eq. $(3)$. Of course, $|S_1(f)|=|S_2(f)|$ implies that the two spectra are related by an all-pass transformation, but the phase of that all-pass can be dependent on the input signal, and hence need not be LTI.

One example of two signals satisfying $|s_1(t)|=|s_2(t)|$ and $|S_1(f)|=|S_2(f)|$ without satisfying $(1)$ is $s_1(t)=e^{j\phi(t)}$ and $s_2(t)=e^{-j\phi(-t)}$. Clearly, $|s_1(t)|=|s_2(t)|=1$. Since $s_2(t)=s_1^*(-t)$, the corresponding spectra satisfy $S_2(f)=S_1^*(f)$, and, consequently, $|S_1(f)|=|S_2(f)|$.

The dual case would be $S_1(f)=e^{j\Phi(f)}$ and $S_2(f)=S_1^*(-f)=e^{-j\Phi(-f)}$. The signals $s_1(t)$ and $s_2(t)$ then satisfy $s_2(t)=s^*_1(t)$ and $|s_1(t)|=|s_2(t)|$.

The second example above requires the signals to be complex-valued (otherwise $s_1(t)=s_2(t)$ would hold). However, the first example also includes the real-valued case, i.e., a signal $s_1(t)$ that only attains two values: either $1$ or $-1$ (like a telegraph process). With $s_2(t)=s_1(-t)$ we guarantee $|s_1(t)|=|s_2(t)|$, and since $S_2(f)=S_1(-f)=S_1^*(f)$ we also have $|S_1(f)|=|S_2(f)|$.


OLD answer:

I think I can show that the trivial case

$$s_2(t)=e^{j\alpha}s_1(t),\qquad\alpha\in\mathbb{R}\tag{1}$$

is the only case for which $|s_1(t)|=|s_2(t)|$ and $|S_1(f)|=|S_2(f)|$ holds.

From the requirement $|s_1(t)|=|s_2(t)|$ it follows that

$$s_2(t)=e^{j\phi(t)}s_1(t)\tag{2}$$

with a real-valued function $\phi(t)$. From the requirement $|S_1(f)|=|S_2(f)|$ it follows that $s_2(t)$ can be obtained from $s_1(t)$ by all-pass filtering:

$$s_2(t)=\int_{-\infty}^{\infty}s_1(\tau)h_{AP}(t-\tau)d\tau\tag{3}$$

with $H_{AP}(f)=\mathcal{F}\{h_{AP}(t)\}$ satisfying $|H_{AP}(f)|=1$.

Since the multiplication in $(2)$ is memoryless, Equations $(2)$ and $(3)$ can only give the same result if $h_{AP}(t)=e^{j\alpha}\delta(t)$, i.e., for a trivial all-pass which just scales the input by a complex constant with magnitude equal to $1$. Consequently, the function $\phi(t)$ in $(2)$ must be constant (and equal to $\alpha$). This means that $(2)$ and $(3)$ are reduced to Eq. $(1)$.

Consequently, given $|s_1(t)|$ and $|S_1(f)|$, the signal $s_1(t)$ is determined up to a complex factor with magnitude $1$. I just do not know of any method to actually compute $s_1(t)$ from these two magnitude functions.