How does the velocity of a tachyon transform under a Lorentz boost? Suppose we only consider motion along the $x$ direction for simplicity. If the velocity of the tachyon is $u$ in the lab frame, what is the velocity $u'$ of the tachyon in a frame moving with velocity $v$ (slower than $c$) relative to the lab frame? Can we still use the formula
$$
u' = \frac{u-v}{1-\frac{uv}{c^2}}
$$
Best Answer
The short answer is that it is valid, but that misses a lot of subtleties.
Using three-vectors for velocity in special relativity is fairly unnatural even for ordinary speeds, and it's worse for tachyons, since their three-velocity can be "infinite" and, in a certain sense, "time reversed" (see below), and those situations can't be described properly with three-vectors.
It's better to use a four-vector to represent speed. The four-velocity at a point is just a tangent vector to the worldline at that point. A three-velocity $(u_x,u_y,u_z)$ is equivalent to an unnormalized four-velocity of $(1,u_x,u_y,u_z)$, or any positive scalar multiple of that.
All four-vectors transform in the same way. Under a Lorentz boost in the $x$ direction by $v, |v|<1$ (using a three-velocity $v$ in anticipation of recovering the formula quoted in the question), that four-velocity becomes $(γ(1 - vu_x), γ(u_x - v1), u_y, u_z)$. You can renormalize that to make the time component equal to $1$ again, getting
$$\left( 1,\; \frac{u_x-v}{1-vu_x},\; \frac{u_y}{γ(1-vu_x)},\; \frac{u_z}{γ(1-vu_x)} \right)$$
and if you drop the $1$, and add explicit factors of $c$, you get a generalization of the formula from the question (which is the special case where $u_y=u_z=0$).
Nothing in that derivation depends on the speed $u$ being light speed or less as such. However, when you renormalize the boosted four-vector, you have to divide by the time component, and when the speed is tachyonic (and only when it's tachyonic), that time component may be zero. In that case, you can represent the speed by a "formally infinite" three-velocity that has an infinite magnitude and a direction like that of any nonzero vector. However, this is mathematically suspect, and really the only reason it makes sense is because it's equivalent to the four-vector form. There is also no natural reason for the velocity to be treated as "infinite" in the first place. Whether a velocity is "infinite" in this sense is frame-dependent and hence not really a property of the velocity.
The time component can also be negative, and because of that, when you divide by it, you lose sign information. (This is when "the tachyon appears to travel backwards" as you observed in a comment.) Whether the sign information matters depends on the nature of the hypothesized tachyons. If tachyons have an arrow of time, and the four-velocity points in the direction of increasing proper time (as one normally imagines it does for sublight speeds), then the direction of the arrow of time is lost in the conversion. However, if tachyons have an arrow of time, then you can use them to send signals into your own past (using a "tachyonic antitelephone"), so perhaps they can't have one and the loss of sign information doesn't matter.
Ultimately, though, the three-velocity is just a poor way of thinking about the velocity of tachyons, even though it can be adapted to that purpose. It's better to stick with the four-velocity for both subluminal and superluminal speeds.