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Fairly straightforward. I suspect that we'd notice the gravitational effects of it long before we'd be able to see it, but at what point in history would people start noticing that stars jitter around in a specific point in the sky every month?
Note: A black hole with an Schwartzchild Radius the size of the Moon would have the mass of 588 Suns. Tidal forces would shred the entire solar system, so that wouldn't make a very good world BUILDING question.
John Dallman calculated that the Moon-mass black hole will have a Schwarzschild radius of ~50 micrometers. That's pretty small, but will the lensing effects make up for it? We can get specific about it by solving for the angular size of the BH's Einstein ring when lensing the disk of the Sun:
$$\theta_{1}=\sqrt{\frac{4GM}{c^{2}}\cdot\frac{D_{LS}}{D_{S}D_{L}}}$$ Where:
m^3 kg^-1 s^-2,kg,m s^-1,
Angular diameter distance is distance defined in terms of an object's physical size, $x$, and its angular size, $\theta$, as viewed from Earth, $d_{A}=\frac{x}{\theta}$. Angular diameter of the Sun is about $0.53\frac{\pi}{180\cdot3600}$ radians. The angular diameter of our black hole however . . . is nearly zero. For small angles, $\theta \approx \frac{x}{d}$, where $x$ is the transverse size of the object and $d$ is our distance from it. Plugging in our values we get: $\frac{5.0\cdot10^{-5}}{3.84\cdot10^{8}}=1.3\cdot10^{-13}$ radians.
Let's see how this pans out . . . Completing the computation, we find the angular diameter of our Einstein ring to be: $\theta_{1}=2.716\cdot10^{-13}$ radians, or about $0.00000008$ arc seconds. The Hubble Space Telescope has an angular resolution of around $0.04$ arc seconds (at 500 nm wavelengths), and the human eye can resolve objects as small as $40$ arc seconds.
I'm not sure if a telescope the size of Earth itself could resolve even that . . .
Edit: Starfish Prime calculated the maximum size of the Einstein ring radius to be $0.15$ arc seconds when the object being lensed is 2x lunar distance from Earth and situated exactly behind the black hole. Maybe a comet on close approach, or a near-Earth asteroid could give such a large flare. The closest comet ever observed was Comet Tempel–Tuttle in 1366 at around ~9x lunar distance.
Nearby objects (up to 2x lunar distance) will have larger lensing effects, but for background objects like the Sun and stars the Einstein ring radius approaches $0.00000008$ arc seconds.
Edit 2: Here's a Desmos calculator of all this to play around with: https://www.desmos.com/calculator/pot7wiymj6
The largest contributing factor to the Einstein ring radius (when all else is constant) is the angular size of the black hole in the sky. A micrometer-scale black hole viewed from hundreds of thousands of kilometers away is really just too small to be seen.
There's an online Hawking Radiation calculator. That tells me that a black hole with the same mass as the moon (0.0123 Earth masses) is only a tenth of a millimetre in diameter and has an effective temperature of only 1.6 degrees Kelvin. The time until it evaporates is extremely long at 5.8E44 years.
Its tidal effects are far more noticeable than its effects on our view of stars; I suspect its presence would not have been deduced until Newtonian gravity was understood, and that might well have been delayed by the absence of a visible Moon.
So, this is slightly awkward to calculate, but I think I've found a nice simple answer that doesn't require anyone to integrate anything, which is nice.
The effect you're interested in is called gravitational lensing whereby light from a distant object is bent around a massive object between the emitter and the observer.
We know the mass of the moon $M$ (~7.4x1022kg), and how far away it is $D$ (~385000 km). The lensing effect is at its strongest when the lens is exactly mid way between the emitter and the observer, and the strength of the lensing effect can be shown by the Einstein ring radius:
$$\theta_e = \frac{4GM}{Dc^2}$$
where $G$ is the gravitational constant and $c$ is the speed of light.
$\theta_e$ turns out to be ~0.15 seconds of arc.
That's quite small. The smallest feature that can theoretically be resolved by a telescope has an angle of $\theta \approx 1.22\frac{\lambda}{d}$ (see Airy disc or diffraction limit) where $\lambda$ is the wavelength of the light you're considering and $d$ is the aperture of your telescope. In theory, then, a perfect telescope with an aperture diameter of ~80cm observing 500nm light should do the job... unfortunately, atmospheric interference limits resolving power of terrestrial telescopes to no smaller than about 0.3-0.5 seconds of arc, even for great big telescopes on the top of high mountains.
This means that until the advent of fancy toys like space-based telescope (eg. Hubble) or clever interferometry (like the pair of telescopes at Keck observatory) the fuzzy blob that your moon-lens generates is simply indistinguishable from any other fuzzy blob your telescopes can see.
Hubble was launched in 1990, and the Keck interferometer was fired up in 2001. There's not much scope for spotting lensing earlier than that, I suspect, though maybe the sheer weirdness of the invisible tidal source woudl encourage earlier investigation.
Maybe no Telescope?
Other answers point out the black hole is smaller than a grain of rice. Even if the hole was directly in front of the Sun, you would not see the black spot we are all familiar with.
However! The moon hole's gravity still pulls at things hundreds of thousands of miles away. This is now you see the hole. You watch as it lenses things behind it by bending the gravity around it.
Here is a GIF from the N.A.S.A of lensing when one star passes in front of another. Even though the back star is blocked, some of the light from it bends around and the total brightness is more than the front star alone.
Your moon hole does the same thing. For example when it goes near the rim of the sun, it will appear to create a little lump on the circumference, as light that would otherwise shoot out into space gets bent around and hits the planet. Note the black dot in the picture will not be visible. It is there to indicate the hole's position.
How big of a telescope you need will depend on how big is the lump. If anyone would like to do the calculations for lump size be my guest!
