How to find the general solution of $x^2y'' - x(x+2)y' + (x+2)y = 0$ $$x^2y'' - x(x+2)y' + (x+2)y = 0$$ 
where a particular solution is $y_1(x) = x$
So, can I start off by subbing in $x$ for $y$ since I have a particular solution?
Which would give me:
$\frac{x}{x+2}y'' - y' + y = 0$
If this is a valid approach, then solving this DE would not be too difficult.

I'm not sure how you got that result, since substituting $y=x$ would only give $0=0$, due to it being a solution of the ODE.
What you need to do is to apply reduction of order. That is, to look for a solution of the form
$$ y(x) = y_1(x)v(x) = xv $$
This gives the equation in $v(x)$
$$ x^2(2v' + xv'') - x(x+2)(v + xv') + (x+2)xv = 0 $$
Simplifying we get
$$ v'' - v' = 0 $$
or
$$ v(x) = c_1 + c_2e^x $$
Therefore
$$ y(x) = c_1x + c_2xe^x $$
