Every non-identity element of $SO(3)$ has two fix points as it acts on the unit sphere, which are the axis of the rotation. Two elements commute if they have the same two fixed points - that is, if they rotate along the same axis, or if their axes are orthogonal and they are both $180$ degree turns.

This can be simplified to include the identity element by saying that for $g,h\in SO(3)$, $gh=hg$ if and only if either $$\exists x\in S^2: gx=x=hx$$ or $$\exists x,y\in S^2: x\cdot y=0: hx=x, gy=y, hy=-y, gx=-x$$

A quick outline of the proof.

First you need to show that every non-identity element of $SO(3)$ has exactly two fixed points, necessarily antipodal.

Second if $g,h\in SO(3)$ and $gh=hg$. Let $x$ be a fixed point of $h$. Then $$gx=g(hx)=(gh)x=(hg)x=h(gx)$$ So $gx$ is also a fixed point of $h$. Which means that $gx=x$ or $gx=-x$. If $gx=x$ we are done, so assume $gx=-x$, and let $y$ be a fixed point of $g$. Then we get: $$y\cdot x = (gy)\cdot(gx) = y\cdot(-x)=-y\cdot x$$ (The first equality is the definition of $SO(3)$.) By symmetric, if $h$ doesn't fix $y$, then $hy=-y$.

So we see that $h$ and $g$ either must share a fixed point, or they must both be $180$ degree rotations around orthogonal axes.

This makes the representative examples:

$$g=\left(\begin{matrix}1&0&0\\ 0&\cos\alpha&-\sin\alpha\\ 0&\sin\alpha&\cos\alpha \end{matrix}\right),h=\left(\begin{matrix}1&0&0\\ 0&\cos\beta&-\sin\beta\\ 0&\sin\beta&\cos\beta \end{matrix}\right)$$ and $$g=\left(\begin{matrix}1&0&0\\ 0&-1&0\\ 0&0&-1 \end{matrix}\right),h=\left(\begin{matrix}-1&0&0\\ 0&1&0\\ 0&0&-1 \end{matrix}\right)$$

with all other pairs being group conjugates of such pairs.

That is different from your question about the center of $GL_2(\mathbb C)$ because that question is not about whether two elements commute, but whether a single element commutes with all other elements of the group.