Why does $\phi \to e^{-i\alpha(x)}\phi$ under a gauge transformation $A_{\mu} \to A_{\mu} + \partial_{\mu}\alpha(x)$? In scalar QED, under a gauge transformation $$A_{\mu}\to A_{\mu}+\partial_{\mu}\alpha(x)$$
$\phi$ "can transform as"
$$\phi \to e^{-i\alpha(x)}\phi$$
(source: Schwartz's QFT 8.49). Why is this true? Why does a gauge transformation in $A_{\mu}$ do anything at all to $\phi$, which is a different field? 
Or are we just saying we can construct a theory such that this holds, so let's just declare this to be true and see what happens? (What would scalar QED be like if the $\phi$ field didn't gauge-transform at all? )
 A: 
Are we just saying we can construct a theory such that this holds, so let's just declare this to be true and see what happens?

Yes, this is basically the case. But the assumption is not as arbitrary as it might seem at first. It's not like people went around trying various transformations like $\phi \to \alpha(x) \phi$, $\psi \to \phi \sin \alpha(x)$, etc. until they just happened to stumble across the transformation $\phi \to e^{-i \alpha(x)} \phi$ that corresponded to the correct phenomenology.
Gauge transformations are funny things, because unlike (most) global transformations, they're not actual physical processes. You can't push a button on your experimental apparatus and cause the system you're studying to undergo a gauge transformation, whose effects you can then study empirically. So asking how a physical quantity transforms under gauge transformations is not really an empirical question.
The unifying principle that motivates people to study the particular simultaneous transformations $A_\mu \to A_\mu + \partial_\mu \alpha(x)$ and $\phi \to e^{-i \alpha(x)}$ is that this pair of transformations happens to leave the scalar QED Lagrangian density
$$\mathcal{L} = -\frac{1}{4} F_{\mu\nu}F^{\mu\nu} -\frac{1}{2} (\partial_\mu - i e A_\mu)\phi(x)\ (\partial^\mu - i e A^\mu)\phi(x)$$
invariant for any choice of function $\alpha(x)$. [This particular form of the Lagrangian wasn't just discovered by trial and error either, but by "gauging" a theory with a global symmetry but no gauge fields via "minimal coupling" - simply replacing all partial derivatives on the charged matter fields in the Lagrangian with "covariant derivatives". That's another story.]
On paper, this is just an interesting mathematical quirk of this particular Lagrangian. But empirically, the Lagrangians for all of the fields in the Standard Model have these special forms, where they are left invariant under these transformations that depend on an arbitrary function of spacetime. Moreoever, these "local symmetries" are necessary both for (a) explaining many relations between terms in those Lagrangians (including why certain renormalizable and otherwise-unobjectionable terms are missing entirely), and (b) giving (semi-heuristic) explanations for how to quantize these classical theories.
So when we ask "how does a particular field transform under gauge transformations?", we really mean "how must it transform in order to leave the Lagrangian density invariant under this formal transformation?" If you just considered the transformation $A_\mu \to A_\mu + \partial_\mu \alpha(x)$ without changing $\phi$ at all, then the Lagrangian would not be invariant, because the product rule from the partial derivatives on $\phi$ would bring out extra terms. So this wouldn't be a gauge transformation at all, and it would be meaningless to say that $A_\mu$ "transforms" that way - such a statement would have neither physical nor mathematical content.
You could imagine a different theory with Lagrangian
$$\mathcal{L} = -\frac{1}{4} F_{\mu\nu}F^{\mu\nu} -\frac{1}{2} \partial_\mu \phi(x) \partial^\mu \phi(x).$$
Such a theory would indeed be invariant under the transformation you propose, and would technically be a gauge theory (different from QED). But in this theory the $\phi$ and $A_\mu$ fields would be completely independent and would not interact at all. Since your experimental apparatus is presumably made of matter fields $\phi$, it would be completely unable to detect the gauge field $A_\mu$ in any way, and you could completely ignore its existence.
A: "Or are we just saying we can construct a theory such that this holds, so let's just declare this to be true and see what happens?"
This is precisely what happens. The essence of scalar QED is that it's a theory of a photon $A_\mu$ and a scalar $\phi$ which both transform nontrivially under the same local $U(1)$ symmetry - meaning that the group action is specified by some function $\alpha(x)$. The textbooks show that this essentially uniquely specifies the theory, at least the terms with dimension $\leq 4$ (it's easy to write down interactions that are irrelevant under RG flows).
Your second question is more interesting: suppose that $\phi$ transformed trivially ($\phi$ has charge 0), could you still write down something interesting? The answer is yes. Our inspiration is as follows: if $J_\mu$ is a conserved current, then
$$\int\!d^4x\,  A^\mu J_\mu$$
is gauge-invariant. This is easy to check, since under the $A_\mu$ variation it reads
$$\int\!d^4x\,  (\partial_\mu \alpha(x))J_\mu = -\int\!d^4x\,  \alpha(x)\partial^\mu J_\mu = 0.$$
Given a complex boson $\phi$ with only a mass term, you can construct the vector operator
$$J_\mu = i (\phi \partial_\mu \bar{\phi} - \bar\phi \partial_\mu {\phi})$$
which you can prove is Hermitian and conserved. This is of course the $U(1)$ Noether current, but we don't care about that right now. So you could study the action
$$L = - (F_{\mu \nu})^2 + |\partial_\mu \phi|^2 + m^2 |\phi|^2 + g A^\mu J_\mu$$
which we just proved to be invariant under
$$A_\mu \to A_\mu + \partial_\mu \alpha(x)\,,
\quad
\phi \to \phi\,,
\quad
\bar\phi \to \bar\phi\,.$$ 
This action is very similar to that of scalar QED, up to a term $A_\mu^2 |\phi|^2$. The reason is that $J_\mu$ is no longer well-defined if $\phi$ transforms to $\exp(-i\alpha(x)) \phi$, so there's a conspiracy between the kinetic term, the $J^\mu A_\mu$ term and the $A^2 |\phi|^2$ term to make the whole thing gauge invariant. That's the beauty of scalar QED.
