First, some metamathematical remarks. I admire Mark's patience in explaining the effect of adding a constant on the range of a function, but I find the whole first page hard to follow.
MarkFL said:
Adding a constant to a function has the effect of shifting it vertically, thereby affecting its range. The range of $g(x)=e^x+c$ is therefore $(c,\infty)$. So then what would the domain of $g^{-1}(x)$ be, and how do we shift the domain of a function?
I am not sure it is necessary to talk about the relationship between the domains of $f^{-1}$ and $g^{-1}$ because if the domain of some function $u$ is $D$ and the domain of some $v$ is $\{x+c\mid x\in D\}$, it obviously does not follow that $v(x)=u(x-c)$.
MarkFL said:
We have observed that the graph of $g(x)$ is shifted vertically from that of $f(x)$, hence the graph of $g^{-1}(x)$ must be horizontal shifted (by the same amount) from that of $f^{-1}(x)$. So how can we state this mathematically?
I am not sure it is helpful to talk about graph shifting, either. I am pretty sure that if the OP did not know how adding a constant changes the range, s/he would be lost by the fact quoted above. Horizontal graph shifting is counterintuitive because adding a positive constant to the argument leads to the graph's shift to the left rather than to the right.
And it has been pointed out that instead of monotonicity of $g(x)$ we must show its injectivity.
I think it is much easier to focus on algebra involving concrete numbers $x$ and $y$ instead of talking about domains or graphs as a whole.
So, we have $g(x)=f(x)+c$ and we need to find $g^{-1}$. This means we need to solve the equation $f(x)+c=y$ for $x$. It is solved in the usual way: we look at operations applied to $x$ in the left-hand side and reverse them. There are two operations applied to $x$: first $f$ and then adding $c$. Therefore, we first have to subtract $c$ from both sides to neutralize addition, and then ... (use the fact that $f^{-1}$ is known).