Is $O(2)$ really not isomorphic to $SO(2)\times \{-1,1\}$?

While this isn't true, it's close enough to true that the confusion is quite believable. In fact, for $n$ odd the map $SO(n)\times \mathbb{Z}_2\to O(n)$ given by $(A,m)\mapsto mA$ is an isomorphism. It's clear that this is a bijection, since in odd dimensions the map $A\mapsto -A$ reverses orientation/determinant. And it's a group homomorphism: $(A,m)(A',m')\mapsto mAm'A'=mm'AA',$ the image of $(AA',mm')$. This basically is just because$-SO(n)$ is a coset $SO(n)*-I$ of a central element $-I$.

But for $n$ even this can't happen: central elements are scalar matrices, and scalar matrices of even dimension have nonnegative determinant. We do still get a bijection (even a diffeomorphism) $SO(n)\times \mathbb{Z}_2\to O(n)$, but here instead of sending $(A,-1)$ to $-A$ we must send it, among other possible choices, to $A$ with the sign of its first column flipped: this exhibits $-SO(n)$ as the coset of the block matrix $\begin{pmatrix} -1&0\\0&I_{n-1}\end{pmatrix}=B$. But again, since $B$ is not central, this is not a group homomorphism.

As Qiaochu points out, your proposed map isn't an isomorphism.

But these two groups are nonisomorphic for the simple reason that $SO(2) \times \{-1, 1\}$ is abelian (it is a product of abelian groups) whereas $O(2)$ is not: For example, the orthogonal matrices $$\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}, \begin{pmatrix}-1 & 0 \\ 0 & 1\end{pmatrix} \in O(2)$$ do not commute.

You can think of this perhaps usefully as an infinite version of the fact that $\mathbb{Z}_n \times \mathbb{Z}_2$ is not isomorphic to $D_{2n}$ (for large enough $n$). Put another way, all rotations of the plane commute, but in general reflections of the plane do not.