Let's take care of the routine calculus for you, so you can get to the heart of the problem and enjoy formulating a solution. It comes down to constructing rectangles as unions and differences of triangles.
First, choose values of $a$ and $b$ that make the details as simple as possible. I like $a=0,b=1$: the univariate density of any component of $X=(X_1,X_2,\ldots,X_n)$ is just the indicator function of the interval $[0,1]$.
Let's find the distribution function $F$ of $(Y_1,Y_n)$. By definition, for any real numbers $y_1 \le y_n$ this is
$$F(y_1,y_n) = \Pr(Y_1\le y_1\text{ and } Y_n \le y_n).\tag{1}$$
The values of $F$ are obviously $0$ or $1$ in case any of $y_1$ or $y_n$ is outside the interval $[a,b] = [0,1]$, so let's assume they're both in this interval. (Let's also assume $n\ge 2$ to avoid discussing trivialities.) In this case the event $(1)$ can be described in terms of the original variables $X=(X_1,X_2,\ldots,X_n)$ as "at least one of the $X_i$ is less than or equal to $y_1$ and none of the $X_i$ exceed $y_n$." Equivalently, all the $X_i$ lie in $[0,y_n]$ but it is not the case that all of them lie in $(y_1,y_n]$.
Because the $X_i$ are independent, their probabilities multiply and give $(y_n-0)^n = y_n^n$ and $(y_n-y_1)^n$, respectively, for these two events just mentioned. Thus,
$$F(y_1,y_n) = y_n^n - (y_n-y_1)^n.$$
The density $f$ is the mixed partial derivative of $F$,
$$f(y_1,y_n) = \frac{\partial^2 F}{\partial y_1 \partial y_n}(y_1,y_n) = n(n-1)(y_n-y_1)^{n-2}.$$
The general case for $(a,b)$ scales the variables by the factor $b-a$ and shifts the location by $a$. Thus, for $a \lt y_1 \le y_n \lt b$,
$$F(y_1,y_n; a,b) = \left(\left(\frac{y_n-a}{b-a}\right)^n - \left(\frac{y_n-a}{b-a} - \frac{y_1-a}{b-a}\right)^n\right) = \frac{(y_n-a)^n - (y_n-y_1)^n}{(b-a)^n}.$$
Differentiating as before we obtain
$$f(y_1,y_n; a,b) = \frac{n(n-1)}{(b-a)^n}(y_n-y_1)^{n-2}.$$
Consider the definition of completeness. Let $g$ be any measurable function of two real variables. By definition,
$$\eqalign{E[g(Y_1,Y_n)] &= \int_{y_1}^b\int_a^b g(y_1,y_n) f(y_1,y_n)dy_1dy_n\\
&\propto\int_{y_1}^b\int_a^b g(y_1,y_n) (y_n-y_1)^{n-2} dy_1dy_n.\tag{2}
}$$
We need to show that when this expectation is zero for all $(a,b)$, then it's certain that $g=0$ for any $(a,b)$.
Here's your hint. Let $h:\mathbb{R}^2\to \mathbb{R}$ be any measurable function. I would like to express it in the form suggested by $(2)$ as $h(x,y)=g(x,y)(y-x)^{n-2}$. To do that, obviously we must divide $h$ by $(y-x)^{n-2}$. Unfortunately, for $n\gt 2$ this isn't defined whenever $y-x$. The key is that this set has measure zero so we can neglect it.
Accordingly, given any measurable $h$, define
$$g(x,y) = \left\{\matrix{h(x,y)/(y-x)^{n-2} & x \ne y \\ 0 & x=y}\right.$$
Then $(2)$ becomes
$$\int_{y_1}^b\int_a^b h(y_1,y_n) dy_1dy_n \propto E[g(Y_1,Y_n)].\tag{3}$$
(When the task is showing that something is zero, we may ignore nonzero constants of proportionality. Here, I have dropped $n(n-1)/(b-a)^{n-2}$ from the left hand side.)
This is an integral over a right triangle with hypotenuse extending from $(a,a)$ to $(b,b)$ and vertex at $(a,b)$. Let's denote such a triangle $\Delta(a,b)$.
Ergo, what you need to show is that if the integral of an arbitrary measurable function $h$ over all triangles $\Delta(a,b)$ is zero, then for any $a\lt b$, $h(x,y)=0$ (almost surely) for all $(x,y)\in \Delta(a,b)$.
Although it might seem we haven't gotten any further, consider any rectangle $[u_1,u_2]\times [v_1,v_2]$ wholly contained in the half-plane $y \gt x$. It can be expressed in terms of triangles:
$$[u_1,u_2]\times [v_1,v_2] = \Delta(u_1,v_2) \setminus\left(\Delta(u_1,v_1) \cup \Delta(u_2,v_2)\right)\cup \Delta(u_2,v_1).$$
In this figure, the rectangle is what is left over from the big triangle when we remove the overlapping red and green triangles (which double counts their brown intersection) and then replace their intersection.
Consequently, you may immediately deduce that the integral of $h$ over all such rectangles is zero. It remains only to show that $h(x,y)$ must be zero (apart from its values on some set of measure zero) whenever $y \gt x$. The proof of this (intuitively clear) assertion depends on what approach you want to take to the definition of integration.
Best Answer
The joint distribution off all order statistics $X_{(1)}, \dots, X_{(n)}$ is $$f_o(y_1, \dots, y_n) = n! f(y_1) \times \cdots \times f(y_n)$$ for $y_1 \le \cdots \le y_n$.
Thus, the joint distribution of $X_1 ,\dots, X_n$, given $X_{(1)}, \dots, X_{(n)}$ does not depend of the density $f$! We have $$ Pr(X_1 = x_1, \dots, X_n= x_n | X_{(1)} = y_1, \dots, X_{(n)} = y_n ) = {1\over n!} $$ when the multisets of the $x_i$'s and of the $y_i$'s are equal.
The intuition is that loosing the drawing order doesn’t matter, for the $X_i$'s are independent.
PS Re-reading this answer long after, I tend to think that the conclusion is clear in itself: given $y_1 < \dots < y_n$, $Pr(X_1 = x_1, \dots, X_n= x_n | X_{(1)} = y_1, \dots, X_{(n)} = y_n ) = {1\over n!}$ means that all re-orderings of the $y_i$'s are equally probable, whatever the density $f$ is. In fact I don't see how to prove the starting result I used (the joint distribution of order statistics) without proving this at the same time... This just comes from the fact that all the points $x = (x_1, \dots, x_n)$ obtained from a permutation of the $y_i$ have the same density $\prod_i f(x_i) = \prod_i f(y_i)$.