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.