Chandru1: Here is a hint to get you started: If the equation $(I-(AB-BA))^n=0$ holds, then for any $v$, if $w=(I-(AB-BA))^{n-1}v$, then $(I-(AB-BA))w=0$, or $(AB-BA)w=w$. If $w=0$ for all $v$, then $(I-(AB-BA))^{n-1}=0$. Proceeding inductively this way, you get that either $AB-BA=I$, or else 1 is an eigenvalue of $AB-BA$.
Now, the first option is impossible, because the traces of $AB$ and $BA$ coincide, but the trace of $I$ is $n$ rather than 0.
If $A$ is a nonsingular matrix with rows $r_1,r_2,\ldots,r_n$, then $\{r_2,\ldots,r_n\}$ spans an $(n-1)$-dimensional subspace $P$ of $\mathbb R^n$. At least one of the standard basis vectors $e_1,e_2,\ldots,e_n$ is not in $P$, say $e_i$. Then $\{e_i,r_2,r_3,\ldots,r_n\}$ is a basis of $\mathbb R^n$, and it follows that there is a real number $c$ such that $r_1-ce_i$ is in $P$. The matrix $A'$ with rows $(r_1-ce_i),r_2,r_3,\ldots,r_n$ is singular, and it is obtained from $A$ by subtracting $c$ from the entry in the first row and $i^\text{th}$ column.
Here's a way to rephrase this somewhat more geometrically. The subspace $P$ is a hyperplane that divides $\mathbb R^n$ into two half-spaces, and $r_1$ lies in one of these halves. The line through $r_1$ in the direction of a vector $v$ has the form $\{r_1+tv:t\in\mathbb R\}$. This line is parallel to $P$ only if $v$ is in $P$; otherwise, the line will cross $P$. Since $P$ can't be parallel to all of the coordinate directions (or else it would fill up all of $\mathbb R^n$), there must be a line of the form $\{r_1+te_i:t\in\mathbb R\}$ that crosses $P$, where $e_i$ is the standard basis vector with a $1$ in the $i^\text{th}$ position and $0$s elsewhere. This means that there exists $t_0\in \mathbb R$ such that $r_1+t_0e_i\in P$. And then, linear dependence of the vectors $r_1+t_0e_i,r_2,\ldots,r_n$ means that the matrix with those rows is singular.
Best Answer
By multiplying both sides in $A$ we obtain $$I+AB^{-1}=A(A+B)^{-1}=(I+BA^{-1})^{-1}$$by defining $U=AB^{-1}$ we have $$I+U=(I+U^{-1})^{-1}$$therefore $$(I+U)(I+U^{-1})=I$$which leads to $$I+U+U^{-1}=0$$by multiplying both sides in $U(U-I)$ we obtain $$(I+U+U^{-1})U(U-I)=(U^2+U+I)(U-I)=U^3-I=0$$therefore $$U^3=I$$ which means that $$\det(U)=1$$or equivalently$$\det(A)=\det(B)$$
Counter example on $\Bbb C$
Let $A=I_2$ and $b=kI_2$ with $k={-1+i\sqrt 3\over 2}$. Then we have $$A^{-1}+B^{-1}=I+{1\over k}I\\(A+B)^{-1}={1\over 1+k}I$$since $$k^2+k+1=0$$ we have $$A^{-1}+B^{-1}=(A+B)^{-1}$$but $$\det(B)=k^2=-1-k={-1-i\sqrt 3\over 2}\ne 1=\det(A)$$
Comment
Even on $\Bbb C$, from $U^3=I$ we can conclude that$$\left|\det(A)\right|=\left|\det(B)\right|$$