(While I find the other answers pretty good, I think there is a possible solution that is pretty plausible but hasn't been mentioned yet.)
TL;DR: You might want to make your sky brighter and your moon bluer, darker and bigger (the latter three so that its brightness stays the same, while it makes less contrast to the sky).
The reason why some objects aren't visible in daylight from Earth is (roughly) that the sky itself (the atmosphere, concretely) shines by diffusing light from the Sun, outshining most celestial objects. It is said that the Moon has more surface brightness than the sky.
In the following, I'll write about magnitudes, which are an arbitrary measure of brightness. It admits negative values, and the lower the magnitude, the brighter. It is logarithmic with base $2.5$, meaning that a difference in 1 magnitude means one object is $2.5$ times brighter than the other.
An object's apparent magnitude (total brightness) is the result of a surface brightness integrated through its visible solid angle (the apparent area of sky it covers):
$S=m+2.5\log_{10}A$, where:
- $S$ is the object's surface brightness
- $m$ is its apparent magnitude, and
- $A$ is its area in arcseconds2
For the Moon: with an apparent size of about half a degree (30 arcminutes, 1800 arcseconds):
$A=\left(\frac{D}{2}\right)^2\pi=900^2\pi\approx2.5\times10^6$, and $m=-12.7$ for the full phase. So:
$S=-12.7+2.5\log_{10}2.5\times10^6\approx-12.7+2.5\times6.4=3.3$ on average
For the sky: The sky is bright because it scatters about 6% of the light of the Sun. Assuming optimal conditions this means:
- $A=2.67\times10^{11}\,\mathrm{arcsec}^2$, half a sphere
- $m=−26.74-2.5\log_{10}6\%\approx−26.74-2.5\times-1.2\approx-23.7$
$\therefore S\approx-23.7+2.5\log_{10}2.67\times10^{11}\approx-23.7+2.5\times11.42=-23.7+28.6=4.9$ on average.
But it's actually more complicated than that. It is not constant, varying according to color, altitude, humidity, the Sun's elevation and the angular distance of the point in the sky you are looking at. But apparently it can be in a range about $6$ magnitudes wide.
All this means that the Moon is $2.5^{1.6}\approx4.3$ times brighter than the daytime sky on average. In turn, an alien sky could outshine a moon by a number of combined effects that make this ratio lower:
- The atmosphere could scatter more from the star's light, making the sky brighter. I don't know which gases/thickness/density you would need, nor the maximum brightness you could attain, but an absolute limit is 50% up from our 6% considering that half the light is scattered to the space (as a reference, $\frac{50\%}{6\%}=8.67$, or about $2.3$ magnitudes, but you couldn't see the Sun, since the atmosphere would need to scatter all of its light). If it is possible to get to 25%, the Moon and the sky's brightness become equal, making the former (almost) invisible most of the time.
- The moon could have a hue more alike that of the sky (i.e., it could be bluer), diminishing the contrast between both.
- The moon could be as bright as the Earth's, yet fainter in terms of surface brightness. As a reference, a moon twice as wide (be it bigger or closer) would cover four times as much area, it would need 4 times less surface brightness to attain the same apparent magnitude. Now the Moon is already dark (with an albedo=13.6%), but you can get to 1/4 of that, with a few objects in the solar system being even darker. This Moon would also need to be less dense, so that it doesn't make a mess with the planet's rotation and tides.
My advice, to keep it realistic while making sure your moon is outshined by your sky about all day, is to use a combination of these effects: for example, an atmosphere two to three times brighter, a moon nearly matching the tone of the sky, 70% wider and a third as bright.