Change of Variables in a Second Order Linear Homogeneous Differential Equation Consider the differential equation
$$\frac{d}{dx} \left( x \frac{dy(x)}{dx}\right) + \frac{\lambda}{x} y(x) = 0$$
This a Sturm-Liouville problem where $\lambda \in \mathbb{R}$ corresponds to the (eventual) eigenvalues of the SL operator. To solve the differential equation we preform the following change of variables
$$v(x) = \ln (x) \implies \frac{dy(x)}{dx} = \frac{dv(x)}{dx} \frac{dy(v)}{dv} = \frac{1}{x} \frac{dy(v)}{dv}$$
Plugging in and multiplying through by $x \not = 0$ we find that
$$ \frac{d^2 y(v)}{dv^2} + \lambda y(x) = 0$$
The change of variables in the case of the derivative happens naturally in the definition of the derivative and derivation of the chain rule, but how do the variables change in the case of $y(x)$? 
As an example (not related to the exercise at hand) if $y(x) = \chi_{[0,1]}$ (the indicator on the unit interval) then clearly 
$$ y(e) = 0, \ y(v(e)) = y(1) = 1$$
So we cannot have $y(x) = y(v(x))$. However in a suitable domain we could have
$$ y(x) = y(\ln(e^x)) = y(v(e^x))$$
At this point I am convinced that there is some fundamental error in my understanding of the change of variables method.
 A: General Case for Change of Variables
Consider the following linear second order differential equation with variable coefficients
$$ \frac{d^2y}{dx^2} + a \cdot \frac{dy}{dx}+ b \cdot y = 0 \tag{1}$$
Next, assume that $y$ is a composition of two functions as follows
$$y =  u\circ v \tag{2}$$
where $v$ is some given function that describes your change of variables. Then by chain rule you can show that
$$\begin{align}
\frac{dy}{dx} &=  \left[ \frac{du}{dx} \circ v \right] \frac{dv}{dx} \\ 
\frac{d^2y}{dx^2} &= \left[ \frac{d^2u}{dx^2} \circ v \right] \left[ \frac{dv}{dx} \right]^2 + \left[ \frac{du}{dx} \circ v \right] \frac{d^2v}{dx^2}
\end{align} \tag{3}$$
Next, you can put $(2)$ and $(3)$ into $(1)$ to get
$$ \left[ \frac{dv}{dx} \right]^2 \left[ \frac{d^2u}{dx^2} \circ v \right] + \left[ \frac{d^2v}{dx^2} + a \cdot \frac{dv}{dx} \right] \left[ \frac{du}{dx} \circ v \right] + b \cdot \left[ u\circ v \right] =0 \tag{4}$$
and then operating $\circ v^{-1}$ on both sides of Eq. $(4)$ you obtain
$$ \left[ \left[ \frac{dv}{dx} \right]^2 \circ v^{-1} \right]  \frac{d^2u}{dx^2}  + \left[ \left[ \frac{d^2v}{dx^2} + a \cdot \frac{dv}{dx} \right] \circ v^{-1} \right] \frac{du}{dx} + \left[ b \circ v^{-1} \right] \cdot u = 0 \tag{5}$$

Your Example
In your problem, the ODE is
$$ \frac{d^2y}{dx^2} + \frac{1}{x} \frac{dy}{dx}+ \frac{\lambda}{x^2} y = 0 \tag{6}$$
and also we have
$$\begin{align}
a &:= x \to \frac{1}{x} \\
b &:= x \to \frac{\lambda}{x^2} \\
v&:=x \to \ln(x) \\
v^{-1}&:=x \to e^x
\end{align} \tag{7}$$
now, using $(7)$ in $(5)$ you will get
$$ \frac{d^2u}{dx^2} + \lambda \cdot u = 0 \tag{8}$$
Finally, when you solve $(8)$ then you can simply obtain $y$ appealing to $(2)$.
I think your confusion is due to some bad notation specially in the third equation you wrote! :)
