A friend and I are discussing whether this is an ODE:
$$y'(x)+y(-x)=e^x$$
My friend claims it is not because of the $-x$. IMHO, the differential equation can be written as $F(x,y,y')=0$ with:
$$F=y' + y\circ (-\mathrm{Id}) -\exp$$
and is an ODE, hence. Could you please confirm?
Your friend is definitely right! If it was an ODE, you will be able to find $F\colon\mathbb{R}^3\rightarrow\mathbb{R}$ such that $y$ is solution if and only if $F(x,y(x),y'(x))=0$. Which is not the case because of the $y(-x)$.
The function you defined take functions as arguments and not triplets of reals!