According to the following paper of Taylor:
J. L. Taylor, A joint spectrum for several commuting operators, J. Functional Anal. 6(1970), 172-191.
we have
Let $A= \begin{pmatrix}0&1\\1&0\end{pmatrix}$ and $I= \begin{pmatrix}1&0\\0&1\end{pmatrix}$. By this answer, the Harte spectrum of $(I,A)$ is equal to
$$\sigma_H(I,A)=\{(1,1);(1,-1)\}.$$
I want to understand why the taylor spectrum of $(I,A)$ which is denoted $\sigma_T(I,A)$ is equal also to $\{(1,1), (1,-1)\}$.Proof: 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 $\sigma_T(I,A)$. We can write similar formulas, but using $R_A(\mu)$ instead. Hence, we 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 $A$.
For more details about the taylor spectrum in a more general context we have:
Best Answer
Chose a basis so that $A\equiv\begin{pmatrix}1&0\\0&-1\end{pmatrix}$. Now note that $$(I-1)\oplus (A-1)\equiv0\oplus\begin{pmatrix}0&0\\0&-2\end{pmatrix},\qquad (I-1)\oplus (A+1)\equiv0\oplus\begin{pmatrix}2&0\\0&0\end{pmatrix}$$ both fail to be injective maps $\Bbb C^2\to\Bbb C^2\oplus \Bbb C^2$. For that reason you achieve $$\{(1,1),(1,-1)\}=\sigma(I)\times\sigma(A)\subset\sigma_T(I,A).$$ Your question contains the proof that $\sigma(a_1)\times\sigma(a_2)\supset \sigma_T(a_1,a_2)$ in general. You can generalise this specific example to see $\sigma_T(I,a)=\{1\}\times\sigma(a)$, provided that $a$ is bounded. (Here you need to use the theorem that if $a$ is a bijective linear map between Banach spaces it is an isomorphism.)