If we allow transformations of both the independent and the dependent variable, then the statement is tautologically true. Say you have a linear second order ODE for a function y(x). Let $y_1(x)$ and $y_2(x)$ be linearly independent solutions. Define $t=y_2(x)/y_1(x)$, and $z=y/y_1$. In the transformed variables, the general solution is $z=c_1+c_2t$, so the equation is $z''(t)=0$. Of course, this is not useful for solving any equations, since we have to know the solutions to define the transformation.
The statement as quoted focusses on the independent variable, which seems odd to me. If we have any differential equation solvable in terms of elementary functions, and we set $y=z\phi(x)$, where $\phi$ is a known elementary function, we get another equation solvable in terms of elementary functions. However, this is a change of the dependent variable, not the independent variable. Also, Denis rightly points out the that any two functions can be made into solutions of a second order ODE.
If we allow transformations of both the independent and the dependent variable, then the statement is tautologically true. Say you have a linear second order ODE for a function y(x). Let $y_1(x)$ and $y_2(x)$ be linearly independent solutions. Define $t=y_2(x)/y_1(x)$, and $z=y/y_1$. In the transformed variables, the general solution is $z=c_1+c_2t$, so the equation is $z''(t)=0$. Of course, this is not useful for solving any equations, since we have to know the solutions to define the transformation.
The statement as quoted focusses on the independent variable, which seems odd to me. If we have any differential equation solvable in terms of elementary functions, and we set $y=z\phi(x)$, where $\phi$ is a known elementary function, we get another equation solvable in terms of elementary functions. However, this is a change of the dependent variable, not the independent variable. Also, Denis rightly points out the any two functions can be made into solutions of a second order ODE.