= Solution
For $\lambda>0$, define $\widetilde g_t(z)=\lambda^{-1/2}g_{\lambda t}(\sqrt\lambda z)$. Differentiating the <Chordal Loewner equation> gives
$$
\partial_t\widetilde g_t(z)=\frac2{\widetilde g_t(z)-\widetilde W_t},
\qquad \widetilde W_t=\lambda^{-1/2}W_{\lambda t}.
$$
<Brownian scaling> makes $\widetilde W$ have the original driver law. The corresponding hull is $\lambda^{-1/2}K_{\lambda t}$, so
$$
\boxed{(K_{\lambda t})_{t\geq0}\overset d=(\sqrt\lambda K_t)_{t\geq0}.}
$$
The time scaling applies to the whole coupled hull process. In particular, the joint law of swallowing times satisfies $(T(cx),T(cy))\overset d=c^2(T(x),T(y))$.
Thus $F(cx,cy)=F(x,y)$. A pair $0<x<y$ is determined up to dilation by $z=y/(y-x)>1$: dilating by $(y-x)^{-1}$ turns it into $(z-1,z)$. Define $f(z)=F(z-1,z)$. Then
$$
\boxed{F(x,y)=f\!\left(\frac{y}{y-x}\right).}
$$
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