how does addition of identity matrix to a square matrix changes determinant? Suppose there is $n \times n$ matrix $A$. If we form matrix $B = A+I$ where $I$ is $n \times n$ identity matrix, how does $|B|$ - determinant of $B$ - change compared to $|A|$? And what about the case where $B = A - I$?

As others already have pointed out, there is no simple relation. Here is one answer more for the intuition. 
Consider the (restricting) codition, that $A_{n \times n}$ is diagonalizable, then $$\det(A) = \lambda_0 \cdot \lambda_1  \cdot \lambda_2  \cdot \cdots \lambda _{n-1} $$
Now consider you add the identity matrix. The determinant changes to
 $$\det(B) = (\lambda_0+1) \cdot (\lambda_1+1)  \cdot (\lambda_2+1)  \cdot \cdots (\lambda _{n-1} +1)$$
I think it is obvious how irregular the result depends on the given eigenvalues of A. If some $\lambda_k=0$ then $\det(A)=0$ but that zero-factor changes to $(\lambda_k+1)$ and det(B) need not be zero. Other way round - if some factor $\lambda_k=-1$ then the addition by I makes that factor $\lambda_k+1=0$ and the determinant $\det(B)$ becomes zero. If some $0 \gt \lambda_k \gt -1$ then the determinant may change its sign...
So there is no hope to make one single statement about the behave of B related to A - except that where @pritam linked to, or except you would accept a statement like $$\det(A)=e_n(\Lambda_n) \to \det(B)= \sum_{j=0}^n e_j(\Lambda) $$ where $ \Lambda = \{\lambda_k\}_{k=0..n-1} $ and $e_k(\Lambda)$ denotes k'th elementary symmetric polynomial over $\Lambda$... (And this is only for diagonalizable matrices)
