Show that $f(x) = e^x $ is a function. Can I simply say the following? I feel like I'm either over or under-thinking this, but it makes sense to me since a function has one y for every x and a one-to-one function has the same criteria.
Let $e^a = e^b$
then ln $(e^a) = $ ln$(e^b)$ $\Rightarrow a = b$
Therefore, $e^x$ is a one-to-one function. Since this is a type of function, 
$e^x$ is a function.

You proved the relationship is one-to-one, but did not prove it is a function. The usual definition is that a relation between two sets (domain and codomain, here both $\mathbb{R}$) defines a function if any element of the domain is associated with exactly one element of the codomain. Here, the definition $f(x) = e^x$ associates with $x$ in the domain exactly one number $e^x$ in the codomain, hence $f$ is indeed a function.
As an example, consider the relation $g:\mathbb{R} \to \mathbb{R}$ defined by assigning to $x \in \mathbb{R}$ the $y$ which satisfies the equation $x = y^2$. Clearly then, some elements in $\mathbb{R}$ (negative numbers) are not even mapped by $g$, some others (positive numbers) are mapped to two values at the same time (e.g. $9 \mapsto \{3, -3\}$) and only $0$ is mapped uniquely. Hence, $g$ would not be a function.
As another example, consider the relation $h:\mathbb{R} \to \mathbb{R}$ associating $x \mapsto 1/x$. This is clearly not a function either since $h(0)$ is not defined. But if we restrict $h:\mathbb{R}^+ \to\mathbb{R}$, it becomes a function.
