A nice way to do this, is to recognize that you are in effect verifying that complex numbers commute (and that this holds even when they are represented as 2x2 matrices). So consider a complex number
##a_1 + b_1 i##, given as
##\begin{bmatrix}
a_1 & -b_1\\
b_1 & a_1
\end{bmatrix}= a_1
\begin{bmatrix}
1 & 0\\
0 & 1
\end{bmatrix} +
b_1 \begin{bmatrix}
0 & -1\\
1 & 0
\end{bmatrix} = a_1 \mathbf I + b_1 \mathbf i##
now multiply by some other complex number ##a_2 + b_2 i## and see that
##\big(a_1 \mathbf I + b_1 \mathbf i\big)\big(a_2 \mathbf I + b_2 \mathbf i\big) = \big(a_2 \mathbf I + b_2 \mathbf i\big) \big(a_1 \mathbf I + b_1 \mathbf i\big) ##
because ##\mathbf i## commutes with scaled forms of itself and the identity matrix ##\mathbf I## commutes with everything. Since you are talking about rotation matrices, you are constraining yourself to a determinant of 1 here (aka complex numbers on the unit circle).
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edit: cleaned up some table formatting issues based on below hint