(There's a couple of these questions kicking around, but I didn't see anyone give the "two boosted copies" answer. Generically, I'd say that's the right answer, since it gives an actual causality violation.)
In your scenario, the two planets remain a hundred thousand light years apart. The fact is, you won't get any actual causality violations with FTL that way. The trouble comes if the two planets are moving away from each other. So, let's say that your warp drive travels at ten times the speed of light. Except if the two endpoints of the trip are moving, then what does that mean? Ten times the speed of light relative to which end?
Let's say Tralfamadore is moving at a steady 20% of $c$ (the speed of light), away from Earth. (So, Earth is moving at a steady 20% of $c$ away from Tralfamadore.)
If I leave Tralfamadore (in the direction of Earth) and I am travelling at anything less than 20% of $c$ relative to Tralfamadore, then I am still moving away from Earth. I'll never get home.
Let's say instead I am travelling at 60% of $c$ relative to Tralfamadore. I will catch up to Earth. Relative to Earth, how fast am I approaching? You might guess the answer is 40% of $c$, but it's 45.45%.
Generally, the velocity subtraction formula of relativity is: $$w = (u-v)/(1-uv/c^2)$$
Let's say instead I am travelling at 100% of $c$ relative to Tralfamadore. Plug $u=c, v=0.2c$ into the formula and get $w=c$. Relative to Earth, I am approaching at 100% of $c$! The speed of light is the same for everyone.
So finally, let's say instead I am using your warp drive to travel at 1000% of $c$ relative to Tralfamadore. Relative to Earth, I am approaching at -980% of $c$. In Earth's reference frame, I will arrive on Earth before I leave Tralfamadore. Now you may say this in itself isn't a causality violation, because we've applied Earth's calendar to Tralfamadore. And that's true, but I'll make a round trip:
- In the futuristic Earth year of 3000, Tralfamadore is 98,000 light years away, and receding at 20% of $c$. I leave Earth at 1000% of $c$, relative to Earth.
- In Earth year 13000 Tralfamadore is 100,000 light years away, and I catch up to it. I turn around and leave Tralfamadore at 1000% of $c$, relative to Tralfamadore.
- In Earth year 2796, I arrive home.
Earth's calendar certainly applies to Earth, and I arrived home two centuries before I left. No two ways about it, I'm a time traveller!
There is nothing special about ten times the speed of light. Given a warp drive that moves a certain amount faster than light, you can make the above time machine using two endpoints that are moving apart a certain amount slower than light, provided that the warp drive can move faster than light relative to either end. This time machine works for any form of FTL: tachyons, warp drives, wormholes, what have you.
In any curved spacetime we can still talk about local reference frames that are small enough scale we can ignore the curvature. We also can ask if there are closed timelike curves (CTC) which basically is asking whether we can time-travel to our past selves. CTCs are strongly thought to be impossible in reality.
The universe is thought to be spatially flat, but the spacetime as a whole is curved. CTC's are impossible: at each point in spacetime you have an "age of the universe". To be precise, this is maximum path-length (proper time) a geodesic could have between the big-bang singularity and said point. Any time-like or light-like curve is moving in the direction of increasing age of the universe; this is just as strong a concept of "future" and "past" as in flat spacetime.
With a single warp-drive you don't have CTC's. But you can still get CTC's with multiple warp-drives. Suppose you build a warp-drive on Earth and send it out into space. You start with an (almost) flat initial-condition and then generate a strongly curved spacetime (your warp bubble). Starting from a flat spacetime (or for very large scales from the spacetime of the universe), is much more physically realistic than starting from any other spacetime. You have to make your weird and wonderful curvature from an "empty canvas" !
With a warp-bubble, the highly curved spacetime is on a small scale. This allows us to glue two bubble spacetimes together so long as the ships don't get very close to each-other. If we consider two Earths, moving relative to each-other, that each make a warp-drive, we can set up the system to generate CTCs. This is one reason we suspect this to be impossible.
There is another reason to suspect making warp drives is impossible: Geodesics would have to diverge in some region, which is an anti-gravity effect. Neither matter nor light can make anti-gravity (antimatter has positive mass just like matter). The "attractive gravity only" rule is more precisely defined as an energy condition and at least one of these is violated by warp drives. Violating certain energy conditions would make the speed of sound faster than light which also allows for time-travel paradoxes.
In general, no known solution with CTC's is physically realistic. They either involve infinitely large systems that cannot be setup from an "empty canvas" or violations of energy conditions. For example, the Kerr metric concentrates it's energy condition violation in it's singularity. Real black holes are thought to lack this feature and be much deadlier instead.
Best Answer
An important caveat: FTL implies time travel only if the FTL mechanism obeys the principle of relativity (that is if there is no "absolute" speed, only relative speed). All known physical laws obey this principle, and it dates back to Galileo, so this is a reasonable assumption.
For simplicity let's consider an arbitrary FTL communication mechanism, and again for simplicity let's assume it is so fast it is "instantaneous" (we'll relax this constraint later). Suppose Bob and Alice are in deep space, moving apart at 0.87c relative to one another, and each is equipped with identical FTL message devices. At time t=0 they passed one another and synchronized clocks.
Suppose Alice sends a signal to Bob when her clock reads 10 hours. What will Bob's clock read when he receives the signal? The principle of relativity says we can assume Alice is at rest, and Bob is moving at 0.87c, so he experiences time dilation relative to Alice with a factor of 2x. So according to Alice, Bob's clock is showing 5 when Alice's shows 10. Thus Alice knows that when her clock reads 10 Bob's will read 5. Since Alice's FTL message is instantaneous, Bob will receive the FTL signal at 5 o'clock (his time).
But the situation is completely symmetric: we could equally well assume Bob to be at rest and Alice to be moving, so in Bob's coordinates Alice's clock is slow and when his clock reads 10 Alice's will show 5. That both observers see the other's clock as running slowly is the root of the so called twin paradox, which we won't go into here (there are many, many questions and answers about it) but it is an established fact of relativity.
Thus: Alice uses her FTL machine to send a message from her 10 o'clock to Bob's 5 o'clock. Bob can then wait 5 hours and send a message from his 10 o'clock to Alice's 5 o'clock. Thus, Alice can use Bob to send a message back in time to herself!
OK, suppose the FTL isn't instantaneous, just very very fast. This doesn't help much: the message gets from Alice to Bob a bit slower, so instead of Bob's clock showing 5 maybe it'll show 6 or 7, and vice-versa. Backwards in time messages are still possible.
If the message is slow enough though, it'll get to Bob after his 10 o'clock, and the paradox is resolved. In this particular case if the FTL message is slower than 2c then no paradox is possible. But if Bob and Alice are moving apart faster, the time dilation effect is stronger and the FTL has to go slower still to avoid paradoxes. In the limit as Alice and Bob are moving apart close to c, the FTL message has to slow down to c to avoid paradox.