For simplicity, let's assume initially that the support of $\mu$ is the whole real line, so that $F$ is a homeomorphism $ \mathbb{R} \to (0,1) $. Let's denote $b:=F(0)=\mu(-\infty,0]=1-\mu[0,+\infty)\in(0,1)$.
The relevant function is the strictly convex function $\Phi :(0,1)\to\mathbb{R}$ defined as $$\Phi(t):=\int_{b}^t F^{-1}(s)\, ds $$ which is the Legendre transformation of $\int_0^x F(t)dt$. It has minimum value at (Incidentally$t=b$, $\Phi(0)=-\int_{-\infty}^0 F(t)dt$and $\Phi(t)\to+\infty$, $\Phi(1)=\int_0^{+\infty} (1 -F(t))dt$both for $t\to0$ and $t\to1$ $\|\Phi\|_\infty=\max\{\Phi(0),\Phi(1)\}$ may be finite or not(indeed $\Phi(0)=\int_{b}^0 F^{-1}(s)\, ds=\int_ {-\infty}^0 F(t) dt$: writing them in terms of the density function $\mu(s)$ as double integrals, and using Tonelli's theorem, one finds that both integrals diverge).
Let's denote $\Psi_{+}:=(\Phi_{|[b,1)})^{-1}:[0,\Phi(1))\to[b,1)$$\Psi_{+}:=(\Phi_{|[b,1)})^{-1}: \mathbb{R}_+ \to[b,1)$ and $\Psi_-:=(\Phi_{|(0,b]})^{-1}:[0,\Phi(0))\to(0,b]$ (they can be viewed as monotone $C^1$ function on $[0,+\infty)$ extending them if needed to the constant value$\Psi_-:=(\Phi_{|(0,b]})^{-1}:\mathbb{R}_+\to(0,b]$ $1$ resp. $0$). Note that $\Psi(t):=1-\Psi_+(t)+\Psi_-(t)=\big|\{\Phi\ge t\}\big|$ is the inverse function of the monotone decreasing rearrangement of $\Psi$, and is supported on $[0,\|\Phi\|_{\infty}]$.
Consider the solution $u:[0,+\infty)\to \mathbb{R}$ of the first order autonomous ODE
$$u'(t) = \frac{1}{2} \Psi (u ),\quad t\ge0$$ with $u(0)=0$, given explicitly by $$u^{-1}(x)=2\int_0^x\frac{dy}{ \Psi (y) }.$$
This integral is finite for all $0\le x< \|\Phi\|_\infty$$x\ge0$ and diverges for $x\to\|\Phi\|_\infty$$x\to+\infty$, so $u$ is a homeo $[0,+\infty)\to[0,\|\Phi\|_\infty)$$\mathbb{R}_+\to\mathbb{R}_+$ .
Then it is easy to check that $\phi_{\pm}:=F^{-1}\circ\Psi_\pm\circ u $ have domainsare defined $[0,\tau_\pm)$$\mathbb{R}_+$, with respectively $\tau_-:=2\int_0^{\Phi(0)} \frac{dy}{\Psi(y)}$ and $\tau_+:=2\int_0^{\Phi(1)} \frac{dy}{\Psi(y)}$ (so at least one amongdiverges to $\tau_-$ and$\pm\infty$ for $\tau_+$ is infinite$x\to+\infty$, and both are infinite if and only if $\Phi(0)=\Phi(1)$); and solve the initial problem written as $$ 2\phi_+ F'(\phi_+)\phi_+'=2\phi_- F'(\phi_-)\phi_-'=1-F(\phi_+)+F(\phi_-)$$ that we can write equivalently, because $\phi_\pm=F^{-1}\circ F\circ \phi_\pm=\Phi'\circ F\circ \phi_\pm$, in the form $$ 2(\Phi\circ F\circ \phi_+)'=2(\Phi\circ F\circ \phi_-)'=1-\Psi_+\circ\Phi\circ F\circ \phi_+ +\Psi_-\circ\Phi\circ F\circ \phi_- .$$
Conversely, if $\phi_\pm$ solve the initial system, the latter equation shows that $\Phi\circ F\circ \phi_+=\Phi\circ F\circ \phi_-$ because they coincide at $0$ and have the same derivative, and in fact solve the above equation for $u$.
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Rmk. The assumption on the support of $\mu$ is not crucial; if it is not the whole real line, $F$ is constant on some open set, and $F^{-1}$ has jumps and it is only defined as a left inverse of $F$; the integral for $\Phi$ is however well defined.