I am trying to "interpolate" an affine transform […]
I'm suggesting a different approach, which should be well suited to interpolation, although it does not depend on splitting the transformation into separate elementary operations the way your question suggests. Instead, I'd use fractional powers of a matrix, as I'll describe now.
Suppose $A$ is diagonalizable (which will be the case for most transformations), then there exists an orthogonal matrix $P$ and a diagonal matrix $D$ such that $A=P\,D\,P^{-1}$. The entries of $D$ are the eigenvalues of $A$, which I'll call $\lambda_1$ and $\lambda_2$. Now you can define $A$ raised to the $t$-th power like this:
$$
A^t = P\,D^t\,P^{-1} = P\,\begin{pmatrix}\lambda_1^t&0\\0&\lambda_2^t\end{pmatrix}\,P^{-1}
$$
Now if you change $t$ continuously from $0$ to $1$, the matrix $A^t$ will change from identity to the matrix $A$. So this is your interpolation.
One thing you have to be careful about is the fact that the $\lambda_k$ will very likely be a conjugate pair of complex number. You can express them as $\lambda_k=e^{z_k}$, where $z_k=\log\lambda_k$, but the logarithm of a number is only defined up to multiples of $2\pi i$. So in this case, you should make sure that the imaginary part doesn't become too big, namely you want $-\pi\le\operatorname{Im}(z_k)\le\pi$. This ensures that the interpolation will not take more turns for a rotation than actually required. Furthermore, you should maintain $z_2=\bar{z_1}$ to make sure that the interpolating matrices will be real as well. With this choice, $\lambda_k^t=e^{t\cdot z_k}$ is well defined and behaves as you described for the case of rotation.
I've created a proof of concept implementation which you can use to experiment, in order to decide whether this is what you want. Here is a snapshot of the kind of interpolation this will create:
With that experiment, I realized that one should include the translative part into the matrix as well, in just the way you composed a matrix including $A$ and $b$ in your question. Otherwise, the positions of the interpolated frames will depend on the location of the defining triples in the plane, which I consider undesirable.
With cameras $C_1$ and $C_2$ with respective camera matrices $P_1^{\{W\}} = \begin{bmatrix} R_1 & t_1 \\ \mathbf{0} & 1 \end{bmatrix}$ and $P_2^{\{W\}} = \begin{bmatrix} R_2 & t_2 \\ \mathbf{0} & 1 \end{bmatrix}$, where $W$ denotes the world frame, we want to find the transformation matrix $P_1^{\{2\}}$ that is the transformation from $C_1$ to $C_2$. You can just use $P_1^{\{W\}}$ and $P_2^{\{W\}}$ to find this, since you know they are both given in the same frame. The basic process is to transform from $C_1$ to $W$ to $C_2$.
Step 1:
Given a point $q^{\{1\}}$ in $C_1$, the the world coordinate is given by $q^{\{W\}} = t_1 + R_1 q^{\{1\}}$
Step 2:
Given a point $q^{\{W\}}$ in $W$, the $C_2$ coordinate is given by $q^{\{2\}} = R_2^{-1} (q^{\{W\}} - t_2)$
Step 3:
Combine steps 1 and 2. You have
$$ q^{\{2\}} = R_2^{-1} (q^{\{W\}} - t_2) $$
$$ q^{\{2\}} = R_2^{-1} ((t_1 + R_1 q^{\{1\}}) - t_2) $$
$$ q^{\{2\}} = R_2^{-1} (R_1 q^{\{1\}} + t_1 - t_2) $$
$$ q^{\{2\}} = R_2^{-1} R_1 q^{\{1\}} + R_2^{-1} (t1-t2) $$
which you can write as
$$ q^{\{2\}} = P_1^{\{2\}} q^{\{1\}} $$
where
$$
P_1^{\{2\}} = \begin{bmatrix} R_2^{-1} R_1 & R_2^{-1} (t_1 - t_2) \\ \mathbf{0} & 1\end{bmatrix}
$$
If you'd like to simplify with notation a bit, and knowing that since $R_2$ is orthonormal that $R_2^{-1} = R_2^T$, you can write
$$
P_1^{\{2\}} = \begin{bmatrix} R_2^{T} R_1 & t_{12} \\ \mathbf{0} & 1\end{bmatrix}
$$
where $t_{12} = t_1^{\{2\}} - t_2^{\{2\}}$.
Best Answer
Homography cannot decomposed to the from you wrote. For example the translation $t$ and $c_x,c_y$ cannot be determined since only they sum is known.
See page 8: in this paper