The main thing to realize here is what the parenthesized expression means for different ranges in $x$:
$$u=x-\frac{\lfloor x\rfloor}{1!}+\frac{\lfloor x\rfloor^2}{2!}+\cdots$$
For $0\leq x<1$, the expression is simply equal to $x$.
For $1\leq x<2$, $\lfloor x \rfloor=1$. So you can write this as:
$$u=x-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\cdots=x-1+e^{-1}$$
where I recognized the value of $e^{-1}$ without the first term. For $x>2$, you need to reevaluate it again, you get $x-1+e^{-2}$. In general, you get
$$u(x)=x-1+e^{-\lfloor x \rfloor}$$
So on each part of the integration domain, $du=dx$ (constant doesn't make a difference).
However, it is quite unclear what exactly this integral means. It is not a proper Riemann integral in two ways.
First, it is ambiguous what the integral boundaries mean. Are they meant to refer to values of $u$ or $x$? If $x$, then it is quite unconventional notation, as $u$ to $x$ is not one to one (more $x$ have the same $u$).
The integral can be computed, as $du=dx$ almost everywhere, but you need to be careful, because technically, if $u(x)$ is treated as distribution, there is a discrete negative jump at $x=1$ and $x=2$ etc., so you have discrete (delta function) contributons in addition to the continuous integral (derivative of a Heaviside jump function is a delta function):
$${\rm d}u={\rm d}x(1+(e^{-1}-1)\delta(x-1)+\cdots)$$
Fortunately your integral does not encompass any of the jumps so the whole transformation is pointless, and you can simply ignore it:
$$\int_0^{\pi/4} f(x){\rm d}u=\int_0^{\pi/4} f(x){\rm d}x$$
