Scaling of distance versus time in an expanding universe

Yes.

The answer is kind of obvious, but if you do not see this, just note that the FLRW metric is: $$g=-c^2dt^2+a(t)^2dSigma^2_t,$$ where $dSigma^2_t$ is metric induced up to the scale factor on the spacelike hypersurfaces of constant time $t$. This metric is independent of $t$, from which it follows, that if you have a curve living on one of this hypersurfaces and you are shifting it by the flow generated by $partial/partial t$, then the length of the curve resulted from the shift is just scaled through the factor $a(t)$, which is independent of the curve. As distance is supposed to be length of geodesic from one point to the other on given hypersurface, and the metric is invariant by the flow, the geodesics determining the distance of two points shifted by the flow (i.e. comoving) should indeed be also result of shift by the flow.

In components this is easy to see. First note, that comoving objects retain all their spatial coordinates and the only coordinate that changes along their movement is $t$. As the geodesic giving their distance is also comoving, then in the component expression for length $$l=int_gamma a(t)sqrt{g_{tt}dot{gamma}_{t}^2+g_{rr}dot{gamma}_{r}^2+g_{thetatheta}dot{gamma}_{theta}^2+g_{phiphi}dot{gamma}_{phi}^2}dlambda==a(t)int_gamma sqrt{left(dSigma^2_tright)_{rr}dot{gamma}_{r}^2+left(dSigma^2_tright)_{thetatheta}dot{gamma}_{theta}^2+left(dSigma^2_tright)_{phiphi}dot{gamma}_{phi}^2}dlambda$$ nothing changes with $t$ under the integral in the last expression and therefore

$$l_{t_1}=frac{a(t_1)}{a(t_0)}l_{t_0}.$$ This is true for any comoving curve living on the hypersurface.

To clarify $left(dSigma^2_tright)_{rr}$ is rr-component of the metric $dSigma^2_t$ and $dot{gamma}_{r}=dr(lambda)/dlambda,$ i.e. r-component of the tangent vector to the curve. The coordinates $r$, $phi$ and $theta$ having their meaning just like in wiki.

The last thing to note, is that scale factor is independent of spatial coordinates or the curve for which the length is measured. Therefore if you have two curves with lengths $l_{1;t_0}$ and $l_{2;t_0}equiv kl_{1;t_0},$ with $k$ being some constant, you will get: $$l_{2;t}=frac{a(t)}{a(t_0)}l_{2;t_0}=kfrac{a(t)}{a(t_0)}l_{1;t_0}=kl_{1;t}$$