Can the matrices $A$ and $I+A$ have the same determinant?

The first thing that comes to mind: $n$ even, $A=-I/2$.

Your first question has already many answers. So I'll address your second question.

Edit: in view of Marc van Leeuwen's recommended answer and examples, I will precise how I interpret your question. I read: is it possible that $A$ and $I+A$ have the same complex eigenvalues repeated according to their algebraic multiplicities? I also assume you mean $n\geq 1$. With these assumptions, the answer to your second question is no.

Proof 1: we have $$ \mbox{tr}(A+I)-\mbox{tr} (A) =\mbox{tr}(A)+n-\mbox{tr}(A)=n\geq 1. $$ So $A+I$ and $A$ have distinct traces. In particular, they can't have the same eigenvalues.

Proof 2: let $p_A(X)=\det (XI-A)$ be the characteristic polynomial of $A$. The characteristic polynomial of $I+A$ satisfies: $$ p_{I+A}(X)=\det(XI-I-A)=\det((X-1)I-A)=p_A(X-1). $$ If $A$ and $I+A$ have the same eigenvalues, it means that they have the same characteristic polynomial. This is therefore equivalent to $$ p_A(X)=p_A(X-1). $$ If $p_A(\lambda)=0$, $p_A(\lambda-1)=0$. And by induction $p_A(\lambda -k)=0$ for all integer $k\geq 0$. So $p_A$ is a degree $n\geq 1$ polynomial with infinitely many roots. That's impossible.

Proof 3: much better than proofs 1 and 2, actually. It follows immediately from the definition of the spectrum that $\mbox{Spectrum}(A+I)=\mbox{Spectrum}(A)+1$. Taking the maximum of the real parts of these sets, we get

$$ \max\; \mbox{Re} \;\mbox{Spectrum}(A+I)=\max \;\mbox{Re} \;\mbox{Spectrum}(A)+1. $$ So $A$ and $A+I$ can't have the same spectra, not even mentioning multiplicities.

Note: the last argument works also for the real spectrum as soon as it is nonempty, which happens simultaneously for $A$ and $A+I$. It works also more generally in a Banach algebra. And as pointed out by Marc van Leeuwen, from leonbloy's observation, it can simply be summarized to: there is no nonempty finite subset in $\mathbb{R}$ which is invariant under $\lambda\longmapsto \lambda +1$. You can now replace finite by compact, and $\mathbb{R}$ by $\mathbb{C}$, to get the general Banach algebra case.

For the second question,

$${\bf A p} = \lambda {\bf p} \iff {\bf (A + I)\, p } = (\lambda +1) \, {\bf p}$$

This says that the eigenvalues of $A+I$ are the same eigenvalues of $A$ incremented by 1, hence they cannot be the same. (It can happen, of course, than some eigenvalue of $A+I$ is also an eigenvalue of $A$).

(Update) I assumed here that we are considering the "full" set of eigenvalues (in $\mathbb{C}$). If we restrict to real eigenvalues, the answer also applies; but now, as noted in others answers, the sets can coincide iff they are empty.