Formality of Treating Differential Operators as a Variable in Solving Diff Eqs. There is these methods in an Engineering Book of an Indian author where the particular solution of a linear Differential Equations of higher order are treated using differential operator $D$.
Now, the author justified that in the case where the diff eq. are :
$(D^3 - 3D + 2)y = e^x$
For the particular soln simply put:
$y = \frac{e^x}{D³-3D+2}$ where $D=1$
Clearly the denominator will become zero so the author stated another method in that particular case:
Differentiate the denominator with respect to D and multiply the numerator by $x$
$y= x\frac{e^x}{d\frac{(D³-3D+2)}{dD}} $ where $D=1$
So by induction the general formula
for finding the particular soln of that forcing function becomes:
$y = x^n\frac{e^{ax}}{\frac{d^nf(D)}{dD}}$ where $D=a$ until the numerator is not 0 upon substitution
My question is, why is this valid?
Similarly for the particular soln for linear Differential Equation of the forcing function $x^n$
Simply use the negative Binomial Theorem for the $f(D)$:
Ex. $(D^2+D)y = x^2$
Put $y = (D^2+D)^{-1}(x^2)$
By separating $\frac{1}{D}$ and from the binomial Theorem:
$y = \frac{1}{D}(1-D+D^2....)(x^2)$
For $D^n$ and $n > 2 $ the $ x^2$ becomes 0 and the ​answer becomes:
$y = \frac{x²-2x+2}{D}$ and operate by $\frac{1}{D}$
Why is Differential Operators allowed to be treated in these way?
Also, what particular subject or topic can I read to understand things like these?

Question: Why is Differential Operators allowed to be treated in these way? Also, what particular subject or topic can I read to understand things like these?
Answer:
Your question is like asking about fourier or Laplace transform. Let me discuss using Laplace transform:
Define $L(f(x)) = \int f(x) e^{-sx} dt$. It turns out u can recover $f(x)$ from $L(f(x))$. 
Let Laplace transform of $y=f(x)$ be written as $L(y)$.  
A Property of $L$: $L(\frac{dy}{dx}) = sL(y)$ 
Hence the Differential equation can be written as $P(s) L(y) = L(e^x)$ where $P(s)$ is a polynomial in $s$.
Hence $y = L^{-1}(\frac{L(e^x)}{P(s)})$.
Another Property of $L$:
$L^{-1} (Y_1(s) Y_2(s)) = L^{-1}(Y_1(s)) * L^{-1}(Y_2(s))$ where $*$ is convolution.
We have : $y = e^x * L^{-1}(\frac{1}{P(s)})$. 
Let $L^{-1}(\frac{1}{P(s)})=p_1(x)$ 
$y = e^x * p_1 =\int e^{(x-t)} p_1(t) dt = e^x L(p_1)|_{s=1} = e^x (1/(P(s))|_{s=1}$.
This is exactly what the book suggests. You can see that the idea works only if the right hand side is $e^x$.
Hope its clear.
Now coming to ur first question.
let $L^{-1}(P(s)) = p(x)$. 
We need $\frac{1}{x}\frac{dP(s)}{ds} = \frac{1}{x} \int \frac{d(p(x)e^{-sx})}{ds} dx =  -\int p(x)e^{sx} dx  = -P(s)$
