General solution for $\frac{\mathrm{d}^2 y}{\mathrm{d} x^2} = y$? Start with
$$\frac{\mathrm{d}^2 y}{\mathrm{d} x^2} = y$$
then
$$\frac{1}{\mathrm{d} x} \, \mathrm{d} \left(\frac{\mathrm{d} y }{\mathrm{d} x}\right) = y$$
$$\frac{\mathrm{d} y}{\mathrm{d} x} \, \mathrm{d} \left(\frac{\mathrm{d} y }{\mathrm{d} x}\right) = y \, \mathrm{d} y$$
$$\frac{\mathrm{d} y}{\mathrm{d} x}  = \sqrt{y^2 + c}$$
$$\int \frac{1}{\sqrt{y^2+c}} \mathrm{d} y = x + c_1$$
$$ \ln\left(\sqrt{y^2 +c_0} + y\right) = x + c_1$$
$$ \sqrt{y^2 +c_0} + y = e^{x + c_1}$$
This does look trigonometric and exponential as it should but I don't know how to proceed to simplify it.
I actually know the actual solution but I want to prove it from first principles. This means I don't want to use guess and check methods because they can't handle solving problems in general and I don't want to assume the result in order to prove it.

Your approach is a slightly garbled version of a perfectly fine heuristic (which then is easily checked to have produced genuine solutions):
From $y''=y$, get ${d\over dx}(y')=y$ as you do, then, one way or another, multiply through by $dy$ as you do, to get $y'\,dy'=y\,dy$. It is at this point that you made some manipulation errors, because one would (at least heuristically) get ${1\over 2}(y')^2={1\over 2}y^2+C$. Taking $C=0$, we can easily take the square root(s): $y'=\pm y$, then $dy/y=\pm d$, or $\log y=\pm x+C$. Exponentiating, $y=e^{\pm x}$, which gives the two solutions we knew all along.
This kind of manipulation does appear in many older differential equations sources, and does give a way to anticipate the solutions (without having to know the constant-coefficient equation riff, as standard and basic as the latter is).
EDIT: after @Ian's and Willie Wong's comments... Indeed, taking $C=1$, for example, instead of $0$, gives $\int \sqrt{1+y^2}\;dy=x$, which is easily solved by letting $y=(t+t^{-1})/2$ (although knowing such a trick is perhaps already contrary to the spirit?)...
2nd EDIT: and, also, "discovering" the solutions via a non-linear differential equation (as a result of the manipulation) does create hazards that linear equations don't share, e.g., that solutions don't necessarily add... :)
