Is the assumption that the two reference frames be inertial required in the derivation of transformation equations? In the derivation of Galilean transformations the only assumption is that the two frames are moving with some uniform relative velocity $u$. 
Suppose with respect to some inertial frame $O$ the two frames $S$ and $S'$ are moving with the same uniform acceleration $a$.
Let $V$ be the velocity of $S$ w.r.t. $O$. Similarly, let $V'$ be the velocity of $S'$ w.r.t. $O$.
Furthermore, let $V_0' - V_0 = u$ (const.). Then
$$V = V_0 + at$$
$$V' = V_0' + at$$
Then the relative velocity is $V' - V = u$.
This is the only result required in deriving the Galilean transformation. So why do people assume that the reference frames be inertial. (I know the point is so that Newton's laws would be valid, but exclusively in the derivation of the transformation equation is this assumption needed?) The same applies in the derivation of Lorentz transformation.
 A: If they are not inertial frame ,then the frames can be distinguished from one another i.e, we can conduct experiments and the readings will be different for different frames. Therefore in Galilean transformation as well as in special theory of relativity we use inertial frames for the principle of relativity to hold.
A: What do you mean that's the only result required? Are you referring to the assumption that both frames have the same acceleration, that the relative velocity is constant, or what?
That's true if both frames are accelerating at a uniform velocity - your coordinate transformations would be the same even if Newtonian physical laws don't hold. If they have two separate accelerations then the law "$\mathbf{V'}-\mathbf{V}=u$" (where $u$ is independent of time) holds, and your equation for $v'$ in the frame of $S$ will look Galilean.
In the case of two frames with identical accelerations, by measuring the velocity of the other frame and not taking into account any physical observations, yes you can derive the Galilean transformations. But it's easy to construct situations where this is not true. If the accelerations of the frames are different you don't get a galilean transformation, and if the frames have angular velocity (that is, if they are rotating reference frames) things would get even weirder. (In that case I'm not sure what an interesting question to ask would be!)
The special relativistic version of this would be different. One issue is usually phrased as: If you tie a string between two spaceships accelerating uniformly (with the same acceleration) separated by some distance, the string will break. (Bell's Spaceship Paradox) Since in one frame the other spaceship would look as if it accelerated away, clearly you can't have events transform linearly from one frame to the other, so the Lorentz transform won't hold. I don't know if there's some configuration of reference frames and accelerations that would allow this to hold - that would be something to prove, and you can ask it on stackexchange after phrasing the question precisely! (and giving it a shot yourself. I'd phrase it as something like, "Given a frame $S_1$ and $S_2$, with positions $X_1(t)$ and $X_2(t)$, with $X_1(0)=x_2(0)$ in inertial reference frame $O$, what restrictions must be placed on $X_1$ and $X_2$ so that events transform in a Lorentz-like way from one frame to another? More specifically, do they have to have zero second derivative always." [to avoid lack of clarity, when I say "frame" here I mean "instantaneous rest frame".] )
A: I would write a comment but i don't have the privilege so am writing an answer 

"In the derivation of Galilean transformations the only assumption is that the two frames are moving with some uniform relative velocity u." 

Hint: "In the derivation of Galilean transformations the only assumption is that the two frames are moving with respect to each other with some uniform relative velocity u."
Read this sentecse carefully.
PS: Since you have marched towards relativity remember every measurement is taken w.r.t some frame of reference(coordinate system). $S$ and $S^{'}$ are inertial w.r.t each other but are non-inertial w.r.t $O$.
