Call $T_a$ the set of functions such that $f(x+1)-f(x)=e^{ax}$ for all $x\in\Bbb R$, and consider the map $\Phi_a:T_a\to \Bbb R^{[0,1)}$ defined by $\Phi_a(f)=\left.f\right\rvert_{[0,1)}$. $\Phi_a$ is a bijection, and specifically $$[\Phi_a^{-1}(g)](x)=\begin{cases}g(\{x\})+\frac{e^{a\{ x\}}-e^{ax}}{1-e^a}&\text{if }a\ne 0\\ g(\{x\})+\lfloor x\rfloor&\text{if }a=0\end{cases}$$
Where $\{x\}$ is the real number in $[0,1)$ such that $x-\{x\}\in\Bbb Z$, and $\lfloor x\rfloor=x-\{x\}$ is the largest integer smaller or equal to $x$.
It is clear that $\Phi_a$ is injective and that $\left.\Phi_a^{-1}(g)\right\rvert_{[0,1)}=g$, therefore it is only a matter of verifying that $\Phi_a^{-1}(g)(x+1)-\Phi_a^{-1}(g)(x)=e^{ax}$ for all $x$, which is just algebra.
Added: Another approach is to notice that if you set $g(x)=f(x)+\frac{e^{ax}}{1-e^a}$ if $a\ne 0$ and $g(x)=f(x)-1$ if $a=0$, then you obtain that $g(x+1)=g(x)$, i.e. that the solutions are the functions in the form $\text{(1-periodic function)} +\frac{e^{ax}}{e^a-1}$.
Best Answer
No, they aren't the same concept.
An indeterminate equation is an equation for which there are an infinite number of solutions; that is, there is not enough information within the equation itself to solve the equation, were it not for a "given" value at which to evaluate the function.
E.g. $y = x^2$ is satisfied by ordered pairs $(0,0),(1,1),(-2,4),(2,4)\dots (x,x^2)$.
A solution to a system of equations can also be indeterminate: e.g., the system of two equations, each in three variables, say $x$, $y$, and $z$, is indeterminate.
A functional equation is a function which is defined implicitly in terms of a function or in terms of the function at some value. For a more thorough explanation and examples, reread the links you provide in your question.
[Edited:] Seriously, as Qiaochu states, they are very different. In a functional equation, you need to solve for a function . The functional equation may very well also be an indeterminate equation, in that it admits of more than one solutions (all functions). [Thanks to Qiaochu for pointing out my earlier mis-statement of a functional equation; I've since replaced "equation" with "function"].