Lorentz invariance of the integration measure

It's invariant because the Lorentz group is $SO(3,1)$ and the letter "S" stands for "special" which mathematically means the condition $$\det M = +1.$$ But the determinant is exactly the coefficient by which the volume form gets multiplied when the coordinates are Lorentz-transformed: $$ x \to M\cdot x\quad \Rightarrow \quad d^4 x \to \det M \cdot d^4 x $$ (this determinant-based transformation rule may also be derived if one views the volume form as an antisymmetric tensor with 4 indices) so if the determinant is equal to $+1$, the measure doesn't change. Well, $d^4 x$ is usually interpreted as $|d^4 x|$, so it's actually invariant under the whole $O(3,1)$, including the metrices with $\det M =-1$. And the condition $\det M=\pm 1$ (with "OR") follows from the orthogonality condition itself, so the adjective "special" is really unnecessary when we're already focusing on pseudoorthogonal matrices.

$d^3 x$ is Lorentz contracted and $dt$ is Einstein dilated by the same factor, so these factors disappear in the new $d^4 x'$.