 Solution

A member of both groups satisfies $A^{T}A=I$ and $A^{T}JA=J$. Multiplication of the second identity by $A$ on the left, using $A{A}^T=I$, gives $JA=AJ$. A real two-by-two matrix commuting with $J=\mathrm{diag}(1,-1)$ must be diagonal, since the two off-diagonal entries change sign in this equation. Its diagonal entries are $\pm 1$ by the orthogonal matrix identity, and the restriction $a>0$ forces the first to be $1$. Therefore the intersection is the two-element group