Sorry, no spectacular death ray occurs for a number of reasons:
- The focal point is in the wrong spot
- Water absorbs nearly all the energy
- The angular diameter is insufficient
1. The focal point is in the wrong spot
The mass of the moon is about 7.3 × 1022 kilograms. The density of water is 997 kg/m3. Now assuming the entire moon is liquid water (which is itself a dubious assumption), that yields a volume of water of:
$$ {7.3 \times 10^{22}\:\mathrm {kg} \over 997 \:\mathrm{kg/m^3} }
= 7.3 \times 10^{19} \:\mathrm m^3 $$
Or, a sphere with a radius of 2.6 × 106 meters. (For comparison, our current Moon has a radius of 1.7 × 106 meters.)
The focal length of a ball lens is:
$$ f = {RN \over 2(N-1)} $$
where $R$ is the radius, and $N$ the index of refraction. So for our water moon lens,
$$ {(2.6 \times 10^{6}) 1.333 \over 2(1.333 - 1) } = 5.2 \times 10^6 \:\mathrm m$$
The Moon is about 384 × 106 meters away from Earth, so the focal point isn't anywhere near Earth's surface. At worst, we have a hazard to future lunar missions.
Moreover, spherical aberration makes the focus less than perfect.
Drawing to scale:

2. Water absorbs nearly all the energy
OK, so the focus in in the wrong spot, but what if we ignore that?
Pure liquid water is pretty clear at visible wavelengths, with an attenuation coefficient on the order of 10-2 m-1. That means the transmitted light is reduced by a factor of 1/e for every 100 meters of water. But compared to the size of the Moon, 100 meters is basically nothing, so very little light gets through.
Furthermore, the attenuation coefficient is much higher for ultraviolet and infrared wavelengths, where much of the solar energy is.

3. The angular diameter is insufficient
So water is effectively opaque at lunar sizes, so what if we ignore that also? What if the Moon is replaced with some kind of matter which is completely transparent, and somehow has optical properties which allow it to focus all the light from the sun on to a small area on Earth?
At first glance this would be pretty bad: about 1.3 × 1016 watts of solar power hits the Moon, and our less-dense water Moon is a little bigger, so intercepts even more power. Concentrating that power in a small area would be Really Bad.
But it's not possible to build such an optical system, no matter what kind of matter replaces the Moon, without significantly increasing the angular diameter of the Moon.
Pretend you're an ant. In the absence of any kind of lens, the angular diameter of the Sun is pretty small: 0.53°. Although the Sun is incredibly hot, most directions are the relatively cool "not Sun", and so your total energy exposure, integrated over all possible directions, is manageable.
Now some kid parks a magnifying lens over you. It's huge, covering much of the sky, with an angular diameter of maybe 90°. In almost every direction you look, you see the Sun. From your perspective, it's as if someone put about 32,000 more suns in the sky, and now the sky is mostly "Sun". Integrated over all possible directions your energy exposure is huge, and you promptly burst into flames.
The trouble is the angular diameter of the Moon-lens-death-ray isn't much bigger than the angular diameter of the Sun. The ordinary Sun and Moon have approximately the same angular diameter, and replacing the Moon with less-dense water increased the solid angle by a factor of 2.3 with a commensurate increase in radiance. Certainly enough to be noticeably warmer, but not catastrophically so over the short duration of an eclipse.

It's counterintuitive but true. Try to burn an ant with any optical system which to the ant looks no bigger than the Sun. You can't do it: it would violate the conservation of etendue. (Other references: 1 2)
To present a real threat to Earth's surface, the aliens would have to increase the angular diameter of the Moon. That means one or more of:
- moving it closer
- adding more mass
- making it less dense
- making it non-spherical