Let $m = \mu - \nu$. As in Martin Hairer's answer, we first consider the limits $a \to \infty$ and $a \to 0$ to see that $$ m(\{0\}) = m(\{1\}) = 0 $$ (only here we use the fact that $\mu$ and $\nu$ are probability measures, otherwise they can be arbitrary finite measures). Thus, $m$ is a measure on $(0, 1)$.

We now substitute $a = e^{-b}$ and $$ t = \log(-\log x) , \qquad x = \exp(-e^t) . $$ If $M$ is the push-forward of $m$ under $x \mapsto t$, then we find that $$ \int_{-\infty}^\infty \frac{1}{1 + \exp(-e^{t - b})} M(dt) = 0 $$ for every $b \in \mathbb R$. In other words, the convolution of a finite measure $M$ with the bounded function $$ \phi(t) = \frac{1}{1 + \exp(-e^{-t})} $$ is identically zero.

By an appropriate variant of Wiener’s Tauberian theorem, the distributional Fourier transform of $M$ (a continuous function) is equal to zero on the spectrum of $\phi$ (that is, on the support of the distributional Fourier transform of $\phi$). Thus, it remains to prove that the spectrum of $\phi$ is all of $\mathbb R$.

Fortunately, we can evaluate the distributional Fourier transform of $\phi$ explicitly: if I did not make a mistake, we have $$ \int_{-\infty}^\infty \phi(t) e^{-i z t} dt = \frac{3 \pi}{2} \delta_0(z) + \operatorname{P{.}V{.}} (1 - 2^{-1 + i z}) \Gamma(i z) \zeta(i z) $$ in the sense of distributions, and the support of the right-hand side is indeed $\mathbb R$. (Time permitting, I will get back to this answer to check the details.)

Let $m = \mu - \nu$. As in Martin Hairer's answer, we first consider the limits $a \to \infty$ and $a \to 0$ to see that $$ m(\{0\}) = m(\{1\}) = 0 $$ (only here we use the fact that $\mu$ and $\nu$ are probability measures, otherwise they can be arbitrary finite measures). Thus, $m$ is a measure on $(0, 1)$.

We now substitute $a = e^{-b}$ and $$ t = \log(-\log x) , \qquad x = \exp(-e^t) . $$ If $M$ is the push-forward of $m$ under $x \mapsto t$, then we find that $$ \int_{-\infty}^\infty \frac{1}{1 + \exp(-e^{t - b})} M(dt) = 0 $$ for every $b \in \mathbb R$ (this is again similar to what Martin Hairer did in his answer). In other words, the convolution of a finite measure $M$ with the bounded function $$ \phi(t) = \frac{1}{1 + \exp(-e^{-t})} $$ is identically zero.

By an appropriate variant of Wiener’s Tauberian theorem, the distributional Fourier transform of $M$ (a continuous function) is equal to zero on the spectrum of $\phi$ (that is, on the support of the distributional Fourier transform of $\phi$). Thus, it remains to prove that the spectrum of $\phi$ is all of $\mathbb R$.

Note that $\phi - 1 + \tfrac12 \mathbb 1_{(0,\infty)}$ decays exponentially fast at $\pm \infty$, and therefore its Fourier transform is an analytic function. The distributional Fourier transform of $1$ is just the Dirac measure. Finally, the distributional Fourier transform of $\mathbb 1_{(0,\infty)}$ is equal to an analytic function in $\mathbb R \setminus \{0\}$, and it has a singularity at zero. It follows that the distributional Fourier transform $\hat\phi$ of $\phi$ is equal to an analytic function in $\mathbb R \setminus \{0\}$. Since $\phi$ is real-valued, it is easy to see that $\hat\phi$ is not identically equal to zero neither in $(0, \infty)$ nor in $(-\infty, 0)$. We conclude that the support of $\hat\phi$ is indeed all of $\mathbb R$, as claimed.

(In fact, we can evaluate the distributional Fourier transform of $\phi$ explicitly: if I did not make a mistake, we have $$ \int_{-\infty}^\infty \phi(t) e^{-i z t} dt = \frac{3 \pi}{2} \delta_0(z) + \operatorname{P{.}V{.}} (1 - 2^{-1 + i z}) \Gamma(i z) \zeta(i z) $$ in the sense of distributions, and the support of the right-hand side is indeed $\mathbb R$.)