How do we know $\sin$ and $\cos$ are the only solutions to $y'' = -y$? According to Wikipedia, one way of defining the sine and cosine functions is as the solutions to the differential equation $y'' = -y$.
How do we know that sin and cos (and linear combinations of them, to include $y=e^{ix}$) are the only solutions to this equation?
It seems like a clever combination of inverse polynomials or exponentials could do the same.

Starting from $y'' = -y$, we can add $y$ to form
$$y'' + y = 0$$
This is a homogeneous second order linear differential equation which we can simplify by writing the characteristic polynomial as
$$r^2 + 1 = 0$$
or
$$r = \pm i$$
which are distinct roots. The general solution can be written as  
$$y(x) = c_1\sin(x) + c_2\cos(x)$$
This tells us that $y_1 = \sin(x)$ is a solution and so is any constant multiple of it, such as $c_1\sin(x)$. Similarly, $y_2 = \cos(x)$ is another solution (and so is any function of the form $c_2\cos(x)$). So, by the principle of superposition, we can add $c_1\sin(x)$ and $c_2\cos(x)$ to form the general solution.
As any linear combination of $c_1\sin(x)$ + $c_2\cos(x)$ works, it is clear that $\sin(x)$ and $\cos(x)$ cannot be the only solutions.

Principle of Superposition: If $y_1$ and $y_2$ are any two solutions of the homogeneous equation $y′′ + p(x)y′ + q(x)y = 0$. Then any function of the form $y = c_1 y_1 + c_2 y_2$ is also a solution of the equation, for any pair of constants $c_1$ and $c_2$.
Remark:  However, while the general solution of $y′′ + p(x)y′ + q(x)y = 0$ will always be in the form of $c_1 y_1 + c_2 y_2$, where $y_1$ and $y_2$ are some solutions of the equation, the converse is not always true.  Not every pair of solutions $y_1$ and $y_2$ could be used to give a general solution in the form $y = c_1 y_1 + c_2 y_2$.
You claim that $\sin(x)$ and $\cos(x)$ are the only solutions. Instead, you should focus on linear combinations of these:
$$ y_1 = \cos(x)$$
$$ y_2 = \sin(x)$$
$$ y_3 = \sin(x) + \cos(x)$$
$$ y_4 = 2\sin(x) + 3\cos(x)$$
$$ y_5 = \sin(x) + i\cos(x)$$
$$ y_6 = 10\sin(x) - 11i\cos(x)$$
among many others where $y_i = c_1\sin(x) + c_2\cos(x)$ for constant $c_1,c_2$.
