Notice that $S$ is symmetric real then it's diagonalizable over $\Bbb R$ and since it's rank is $2$ then it's similar to
$$\operatorname{diag}(\lambda_1,\lambda_2,0,\ldots,0)$$
where $\lambda_i\ne0$ so $I-S$ is similar to
$$\operatorname{diag}(1-\lambda_1,1-\lambda_2,1,\ldots,1)$$
and then the result follows easily.
The maximal dimension of subspace of real invertible $n\times n$ matrices is given by the Hurwitz-Radon numbers $\rho(n)$, which is defined as follows: if $n=2^{4a+b}c$ where $0\le b\le3$ and $c$ is odd, then $\rho(n)=8a+2^b$. See
J. F. Adams (1962), Vector fields on spheres, Annals of Mathematics, 75(3): 603-632,
J. F. Adams, P. Lax and R. Phillips (1965), On matrices whose real linear combinations are non-singular, Proc. Amer. Math. Soc., 16:318-322,
J. F. Adams, P. Lax and R. Phillips (1966), Corrections to "On matrices whose real linear combinations are non-singular", Proc. Amer. Math. Soc., 17: 945-947.
The result of Adams (1962) essentially says that the maximal number of linearly independent vector fields on the $(n-1)$-dimensional sphere $S^{n-1}\subset\mathbb R^n$ is $\rho(n)-1$. The following presentation briefly explains the connection of these vector fields with invertible matrix subspace:
Rachel Quinlan, Special spaces of matrices, IMS Meeting 2013, NUI Maynooth.
Here are some obvious properties of the Hurwitz-Radon numbers: (a) $\rho(n)=1$ when $n$ is odd, (b) $\rho(n)\le n$ for every $n$, (c) $\rho(n)=n$ iff $n=1,2,4,8$. From (b) and (c), it follows that there is an $n$-dimensional subspace of real $n\times n$ invertible matrices if and only if $n=1,2,4,8$.
A proof of the weaker statement that the maximal dimension is bounded above by $n$ is given in
Zoran Z. Petrović (1999), On nonsingular matrices and Bott periodicity, Publications de l'Institut Mathématique 65(79).85: 97-102.
Clearly, the results of Adams et al. apply to real invertible matrices only. When the matrix space $M_n(\mathbb C)$, the maximal dimension of a subspace of invertible matrices is obviously $1$, as $A-\lambda B$ is singular when $\lambda$ is an eigenvalue of $AB^{-1}$.
Best Answer
Rank one update, reference Matrix Analysis and Aplied Linear Algebra, Carl D. Meyer, page 475:
If $A_{n \times n} $ is nonsingular, and if $\mathbf{c}$ and $\mathbf{d} $ are $n \times 1$ columns, then \begin{equation} \det(\mathbf{I} + \mathbf{c}\mathbf{d}^T) = 1 + \mathbf{d}^T\mathbf{c} \tag{6.2.2} \end{equation} \begin{equation} \det(A + \mathbf{c}\mathbf{d}^T) = \det(A)(1 + \mathbf{d}^T A^{-1}\mathbf{c}) \tag{6.2.3} \end{equation}
So in your case, $A=\mathbf{I}$ and the determinant is $1(1+ t\mathbf{v}^T\mathbf{v})=1+t$
EDIT. Further from the text:
Proof. The proof of (6.2.2) [the previous] follows by applying the product rules (p. 467) to \begin{equation} \pmatrix{\mathbf{I} & \mathbf{0} \\ \mathbf{d}^T & 1}\pmatrix{\mathbf{I} + \mathbf{c}\mathbf{d}^T& \mathbf{c} \\ \mathbf{0} & 1}\pmatrix{\mathbf{I} & \mathbf{0} \\ -\mathbf{d}^T & 1}=\pmatrix{\mathbf{I} & \mathbf{c} \\ \mathbf{0} & 1 + \mathbf{d}^T\mathbf{c}} \end{equation}
To prove (6.2.3) write $A + \mathbf{c}\mathbf{d}^T = A ( \mathbf{I} + A^{-1}\mathbf{c}\mathbf{d}^T)$, and apply the product rule (6.1.15) along with (6.2.2)