The Taylor spectrum happens to be the same $\sigma_T(I,A) = \{(1,1), (1,-1)\}$. Let $R_X(\lambda) = (X-\lambda)^{-1}$ be the resolvent of $X$. You have the identities
\begin{gather*}
\begin{bmatrix} I-\lambda & A-\mu \end{bmatrix}
\begin{bmatrix} R_I(\lambda) \\ 0 \end{bmatrix} = I, \\
\begin{bmatrix} R_I(\lambda) \\ 0 \end{bmatrix}
\begin{bmatrix} I-\lambda & A-\mu \end{bmatrix}
+ \begin{bmatrix} -(A-\mu) \\ I-\lambda \end{bmatrix}
\begin{bmatrix} 0 & R_I(\lambda) \end{bmatrix}
= \begin{bmatrix} I & 0 \\ 0 & I \end{bmatrix} , \\
\begin{bmatrix} 0 & R_I(\lambda) \end{bmatrix}
\begin{bmatrix} -(A-\mu) \\ I-\lambda \end{bmatrix} = I ,
\end{gather*}
whenever the resolvent $R_I(\lambda)$ exists. These identities (in homological algebra, they are known as a contracting homotopy for this complex) imply that, whenever $\lambda$ is not in the spectrum of $I$ (namely, when $R_I(\lambda)$ exists), Taylor's Koszul complex is exact and hence the corresponding value of $(\lambda,\mu)$ does not belong to $\sigma_T(I,A)$. You can write similar formulas, but using $R_A(\mu)$ instead. Hence, you have reduced the calculation to $\sigma_T(I,A) \subseteq \{ (1,\mathbb{C}) \} \cap \{ (\mathbb{C},1), (\mathbb{C},-1) \} = \{ (1,1), (1,-1) \}$. Now it's just a matter of checking that for these values of $(\lambda,\mu)$ the Koszul complex really does fail to be exact, which is easy to see from the known common eigenvectors of $I$ and $A$.
High brow explanation:
A cochain map between two complexes descends to a map in cohomology. For example, the identity cochain map induces the identity map in cohomology. For any complex, a homotopy induces a cochain map from the complex to itself, which happens to descend to the zero map in cohomology. So, if the identity cochain map from a complex to itself is a induced by a homotopy, then it descends to cohomology as both the identity map and the zero map. That is only possible when the cohomology vanishes, meaning that the complex is exact.
Low brow explanation:
Consider a complex of linear operators $D_i$ (meaning of course that $D_{i+1} D_i = 0$). A homotopy $h_i$ is a sequence of linear maps as illustrated in
$$
\require{AMScd}
\begin{CD}
\cdots V_{-1} @>{D_0}>{\stackrel{\dashleftarrow}{h_0}}>
V_0 @>{D_1}>{\stackrel{\dashleftarrow}{h_1}}>
V_{1} \cdots ,
\end{CD}
$$
Defining the operators $N_i = h_{i+1} D_{i+1} + D_i h_i$, it is easy to check that $D_i N_i = N_{i+1} D_{i+1}$, meaning that $N_i$ constitute a cochain map, which is induced by the homotopy $h_i$. If $N_i = I_i$ is equal to the identity map for each $i$, then we call $h_i$ a contracting homotopy.
If you try to solve the equation $D_{i} v = u$, where $D_{i+1} u = 0$. Then a contracting homotopy can be used as follows:
$$
u = I_{i} u = (h_{i+1} D_{i+1} + D_i h_i) u = D_i (h_i u) .
$$
Hence, a solution $v = h_i u$ always exists. This shows that the complex is exact, whenever a contracting homotopy exists.
