Rather than using the more conventional approach of abusing gravitational effects to produce our doomsday, this answer relies on solar radiation and optics. There is one caveat - it requires a very weird moon which, while scientifically possible, must be an artificially created structure.
Approach:
Make the moon a spherical lens. During an 'eclipse', the moon will focus the sun's rays to a single point on the earth's surface. This will cause rapid concentrated heating, leading to drastic weather changes (as well as melting any location unfortunate enough to fall under the focal point).
This will only occur during a perfectly aligned total lunar eclipse; an imperfect alignment will cause the focal beam to miss the earth.
While this is perhaps somewhat outside of the intended scope of the question, it does fit within the spirit of the question - a celestial alignment causing doomsday-like effects.
Other than the composition of the moon, the solar system is similar to ours for the purposes of interplanetary distances and eclipse frequency.
The Lens:
The effective focal length of a lens is:
$$EFL = \frac{nD}{4(n-1)}$$
(source)
Then, let:
$$D = Distance\;from\;earth\;to\;moon = 384,400\,km$$
$$EFL=Diameter\;of\;moon = 3,474\,km$$
Putting this into the equation, this gives an index of refraction of approximately n = 1.0023. We can achieve something close to this by using benzene gas as our refractive material (n = 1.0018). To get a bit closer to this value of n, we can increase the distance to 49,3774 km or decrease the diameter to 2704 km.
This leaves us with a moon comprised of a solid transparent shell filled with benzene gas or similar for our lens.
Note that this means the moon will be much lighter than our moon, so the planet would not likely experience any tides.
Effects:
During an 'eclipse', the moon lens will focus (most of) the sunlight passing through it to a small point on the earth's surface.
At earth's orbit, the power density of sunlight is approximately 1.36 kW/m2 (source).
Given a diameter of 3,474 km (r = 1.737 x 106 m), the cross-sectional area of the moon will be:
$$\pi r ^2 = 9.4787 \times 10^{12}\,m^2 $$
This means that we will have 1.289x1016 Watts passing through the lens.
If we assume a totality/alignment of about 100 minutes (6000 s) (source), this gives a total energy output of about 7.734x1019 Joules over the course of the eclipse.
While the location of the focal point will (rapidly) move across earth's surface during the eclipse, this is still enough energy to cause plenty of damage. For instance, if the focal point spent most of its time over ocean, it would boil away somewhere around 3 x 1016 g of water. Given that a typical hurricane produces about 2.1 x 1016 g of rain (source), this should be sufficient to produce some spectacular storms.
For some more context on how much energy we are seeing at the focal point, I took a look at https://en.wikipedia.org/wiki/Orders_of_magnitude_(energy).
Notably, the focal point outputs an equivalent amount of energy to the Hiroshima bomb every microsecond.