Why does using "long division" to simplify $1/P(D)$ work when solving $P(D)=x^a$? If you have the differential equation $P(D)=x^a$ you can write it as $(1/P(D))(x^a)$ and use "long division", treating $D^a$ like a regular variable $x^a$ to simplify the operator $1/P(D)$ until the $D^a$ term(because further terms are zero if the solution consists of $C_1,C_2x$...$ C_ax^a$ with constants $C_1$ through $C_n$) and then apply it to $x^a$ to get the particular solution. Why does using "long division" work here, given $(1/P(D))(f(x))$ is defined as the particular solution that doesn't have any terms that have an arbitrary constant in front of them in the general solution of $P(D)=f(x)$?

This works because the differentiation operator $D$ is in fact an object that can be "added" and "multiplied" just like real numbers can. The same is true of any Linear operator on a vector space. Note that $P$ is a polynomial with constant coefficients. It doesn't work if the coefficients of $P$ are allowed to be functions of $x$.
Let $T: V \to V$ be a linear operator (or endomorphism) on the vector space $V$. The set $\text{End}(V)$ of all such endomorphisms is an algebra with composition as the multiplication. If we consider the subalgebra $\scr T$ generated by $T$ (i.e., the smallest subalgebra of $\text{End}(V)$ which contains $T$), it consists of all expressions of the form $P(T)$ where $P$ is a polynomial with coefficients in the field of $V$. Though multiplication on $\text{End}(V)$ is not commutative in general, it is when restricted to $\scr T$.
$\scr T$ obeys all the laws of a field except for the existence of multiplicative inverses. Therefore any purely algebraic manipulation that can be performed in a field which does not require the existence of inverses for arbitrary elements will also work in $\scr T$. This includes the Euclidean algorithm that you are using.
To see it directly, let's divide $1$ by $1 + T + T^2$. This proceeds just like the normal Euclidean algorithm, if one considers the higher powers of $T$ to be of decreasing importance.
$$\begin{align}1 &= 1(1 + T + T^2) + (-T - T^2)\\-T - T^2 &= -T(1 + T + T^2) + T^3\\T^3 &= T^3(1 + T + T^2) + (-T^4 - T^5)\\-T^4 - T^5 &= -T^4(1 + T + T^2) + T^6\end{align}$$
Putting it together: $$1 = (1 + T + T^2)(1 - T + T^3 - T^4) + T^6$$
Now suppose that $v$ is a vector for which $T^6v = 0$. Then $$v = (1 + T + T^2)(1 - T + T^3 - T^4)v$$
That is, $v - Tv + T^3v - T^4v$ is a solution for $w$ in the equation $$(1 + T + T^2)w = v$$
When $V$ is the space of 6-times differentiable functions, $T$ is the differentiation operator, and $v = x^5$, we see that a solution to $$y'' + y' + y =  x^5$$ is provided by $$y = x^5 - \frac{d}{dx}x^5 + \frac{d^3}{dx^3}x^5 - \frac{d^4}{dx^4}x^5$$
Note that all of this is strictly algebraic. It does not require limits. As long as $v$ is annihilated by some power of $T$, this trick can be used to find a $w$ that solves $P(T)w = v$. 
When $v$ is never annihilated (e.g., when $T = D$, and $v$ is not a polynomial), then topology and calculus become involved, and the method might fail.
