The key word under which you will find this result in modern books is "Schur complement". Here is a self-contained proof. Assume $I$ and $J$ are $(1,2,\dots,k)$ for some $k$ without loss of generality (you may reorder rows/columns). Let the matrix be
$$
M=\begin{bmatrix}A & B\\\\ C & D\end{bmatrix},
$$
where the blocks $A$ and $D$ are square. Assume for now that $A$ is invertible --- you may treat the general case with a continuity argument. Let $S=D-CA^{-1}B$ be the so-called Schur complement of $A$ in $M$.
You may verify the following identity ("magic wand Schur complement formula")
$$
\begin{bmatrix}A & B\\\\ C & D\end{bmatrix} =
\begin{bmatrix}I & 0\\\\ CA^{-1} & I\end{bmatrix}
\begin{bmatrix}A & 0\\\\ 0 & S\end{bmatrix}
\begin{bmatrix}I & A^{-1}B\\\\ 0 & I\end{bmatrix}. \tag{1}
$$
By taking determinants, $$\det M=\det A \det S. \tag{2}$$ Moreover, if you invert term-by-term the above formula you can see that the (2,2) block of $M^{-1}$ is $S^{-1}$. So your thesis is now (2).
Note that the "magic formula" (1) can be derived via block Gaussian elimination and is much less magic than it looks at first sight.
For the first point, note that by Sylvester's identity
$$\det(I_n + v^TA^{-1}v) = \det(I_n + (v^T)(A^{-1}v)) = \det(I_n+(A^{-1}v)(v^T)) = \det(I_n + A^{-1}vv^T),$$
so
$$\det(A)\det(I_n+v^TA^{-1}v) = \det(A)\det(I_n + A^{-1}vv^T) = \det(A + vv^T).$$
Given $u \in \mathbb{R}^m$ and $v \in \mathbb{R}^n$, their outer product is the $m\times n$ matrix $uv^T$. If $u$ and $v$ are non-zero, then their outer product has rank one. Conversely, a rank one $m\times n$ matrix can be written as the outer product of some non-zero $u$ and $v$.
Now suppose $C$ is an $n\times n$ matrix (not necessarily invertible) and $A$ is a rank one $n\times n$ matrix. By the above discussion, there are $u, v \in \mathbb{R}^n$ such that $A = uv^T$. So
\begin{align*}
\det(I_n + tCA) &= \det(I_n + tCuv^T)\\
&= \det(I_n + (tCu)(v^T))\\
&= \det(I_1 + (v^T)(tCu))\\
&= \det(I_1 + tv^TCu)\\
&= 1 + tv^TCu
\end{align*}
where the last equality follows because $I_1 + tv^TCu$ is a $1\times 1$ matrix.
A function of the form $f(t) = at + b$ is called affine. I'm guessing this is what is meant by affine-linear.
Added Later: Let me add a theorem which further demonstrates the relationship between the rank of a matrix and outer products. In the following statement, I am taking the definition of the rank of a matrix to be the dimension of its column space.
Theorem: An $m\times n$ matrix $A$ has rank $k$ if and only if the minimal number of outer products needed to express $A$ as a sum is $k$ (i.e. $A$ can be written as the sum of $k$ outer products, but not $k - 1$).
Proving this is a really nice exercise in elementary linear algebra. I would hate to rob you of the experience by posting the solution here.
Best Answer
For me, a standard book to look for such things is Prasolov's linear algebra. This is Theorem 1.2.6.1. Here is the Russian edition, but the proof is essentially a formula.