First, I'll write that identity as $\det(I_n + A) = \sum_{J \subseteq\{1,2,\ldots,n\}} \det(A_J)$ for clarity.
You need to get a summation over $2^n$ terms, while you have $2$ terms in inside the determinant; it is therefore useful to successively apply multi-linearity with respect to each of the $n$ columns (this is similar to expanding a product of $n$ binomials using distributivity, giving $2^n$ terms). Applying it for the first column, one gets $\det(I_n+A)=\det(A_0)+\det(A_1)$ where $A_0$ is the result of replacing the first column of $A+I$ by the first column of $I_n$, and $A_1$ is the result of replacing the first column of $A+I$ by the first column of $A$. Repeat this for each column, giving $2^n$ terms which we label by the subset $J$ of column indices for which we have replaced the column by that of $A$ rather than of $I_n$. I will write that term as $\det(A[J])$ where $A[J]$ is the matrix obtained from$~A$ by replacing all columns with index in the complement of $J$ by the corresponding column of$~I_n$, then we have obtained
$$\det(I_n + A) = \sum_{J \subseteq\{1,2,\ldots,n\}} \det(A[J]).$$
It remains to show that $\det(A[J])=\det(A_J)$, the principal minor of $A$ for the rows and columns indexed by$~J$. This means we must remove from $A[J]$ the columns that were taken from $I_n$ and the rows in which those columns have an entry$~1$. But for a given such column, it is immediate that Laplace expansion by that column does precisely that (remove the column and the row where it has a entry$~1$).
The repeated expansion by linearity in each of the columns could of course (and should if we want a formal proof) be formulated in the form of a proof by induction. If one considers the argument used, it becomes clear however that this requires a bit of induction loading: strengthening the statement to be proved for the sole purpose of making the inductive argument possible. The statement will involve an additional parameter $K\subseteq\{1,2,\ldots,n\}$ which in the final application will be taken to be the whole set, and the induction will be by the size of the set$~K$. The statement to prove by induction then becomes the following, using above notation and in addition $I_K$ to stand for the $n\times n$ diagonal matrix whose diagonal entry at position $i$ is $1$ if $i\in K$, and $0$ otherwise:
$$
\det(A+I_K) = \sum_{J\subseteq K} A[J\cup\overline K]
$$
(the bar designates the set complement inside $\{1,2,\ldots,n\}$). The proof by induction is now straightforward. The starting case $K=\emptyset$ holds by definition (the sum has a single term $J=\emptyset$), and for the inductive step with $K\neq\emptyset$ one writes $K=K^-\cup\{k\}$ (with of course $k\notin K^-$), then applies linearity in column $k$ giving $\det(A+I_K)=\det(A+I_{K^-})+\det(A'+I_{K^-})$ where $A'=A[\overline{\{k\}}]$ is obtained from $A$ by replacing column $k$ by that of $I_n$; then by induction the first term $\det(A+I_{K^-})$ gives the sum of terms $\det(A[J\cup\overline{K^-}])$ for those $J\subseteq K$ that contain$~k$, while the second term $\det(A'+I_{K^-})$ gives a similar sum for those $J\subseteq K$ that do not contain$~k$. I'll leave the (simple but somewhat confusing due to the complements taken) details as an exercise, noting just that the proof for the second term uses that $A'[J\cup\{k\}\cup\overline K]=A[J\cup\overline K]$ when $k\notin J$.
