The correct option is S-I True, S-II True.

Let us analyze the given matrix-valued function:

Step 1: Evaluation of Statement-I (S-I)
We need to check if $f(x)\cdot f(y)=f(x+y)$ is true.
Let us find the product of $f(x)$ and $f(y)$:

Performing row-by-column multiplication, we get:

Using standard trigonometric identity formulas:
1. $\cos (x+y)=\cos x\cos y-\sin x\sin y$
2. $\sin (x+y)=\sin x\cos y+\cos x\sin y$
3. $-(\sin x\cos y+\cos x\sin y)=-\sin (x+y)$

Substituting these identities back into the matrix, we get:

Therefore, Statement-I (S-I) is True.

Step 2: Evaluation of Statement-II (S-II)
We need to verify if $f(-x)$ is invertible.
Let us find the matrix $f(-x)$ by substituting $-x$ in place of $x$:

Since $\cos (-x)=\cos x$ and $\sin (-x)=-\sin x$, we have:

Now, let us calculate the determinant of $f(-x)$:
Expanding along the third column:
$$\det (f(-x))=1\cdot \left|\begin{array}{cc}\cos x& \sin x\\ -\sin x& \cos x\end{array}\right|$$

Evaluating this 2×2 determinant:
$$\det (f(-x))={\cos }^{2}x-(-{\sin }^{2}x)={\cos }^{2}x+{\sin }^{2}x=1$$

Since the determinant $\det (f(-x))=1\ne 0$, the matrix $f(-x)$ is non-singular and therefore invertible.
Thus, Statement-II (S-II) is also True.