That is not true. The reason is that for non-normal (bounded or unbounded, that doen's matter here) operators, all the nice equivalent characterisations of the essential spectra of normal/self-adjoint operators are no longer equivalent. For this reason, you cannot define 'the' essential spectrum of a general operator and there is downright 'zoo' of various essential spectra.
For instance, the set $\sigma_d(T)$ you defined in $(1)$ is the correct definition of the discrete spectrum. But its complement in the spectrum
$$
\sigma_{e5}(T):=\sigma(T)\setminus\sigma_{d}(T),
$$
is, first of all, not equal to $\sigma_{ess}(T)$ as you define it above and, secondly, only invariant under commuting compact perturbations but, in general, not invariant under arbitrary compact perturbations.
The largest essential spectrum that is always stable under any compact perturbations is the set $\sigma_{e4}(T) :=\mathbb{C}\setminus \rho_{e4}(T)$, where
$$
\rho_{e4}(T):=\{\lambda \in \mathbb{C} \mid T-\lambda\ \text{is Fredholm with index }\ 0\}.
$$
However, in general only the inclusion $\sigma_{e4}(T)\subset \sigma_{e5}(T)$ holds which can be strict. An exmple which shows that this can be strict and that $\sigma_{e5}(T)$ is not invariant under arbitrary comapct perturbation can be constructed using the bilateral shift operator.
The essential spectrum defined in Kato is again differtent and can be defined as
$$
\sigma_{e1}(T):=\{\lambda \in \mathbb{C} \mid T-\lambda\ \text{is not semi-Fredholm }\}.
$$
It is even smaller and satisfies
$$
\sigma_{e1}(T) \subset \sigma_{e4}(T) \subset \sigma_{e5}(T),
$$
again, strict inclusions being possbible. As my enumeration suggests, there are also two essential spectra in between (and even much more exists in literatur). Among all the essential spectra that can be defined with Fredholm properties, $\sigma_{e1}(T)$ is the smallest one and $\sigma_{e5}(T)$ is the largest one and they all coincide in the normal case (however, I've also seen an essential spectrum that is even smaller than $\sigma_{e1}(T)$, but isn't defined in terms of Fredhol properties).
If you would like or need to know more about the various essential spectra, I would recoomend having a look into Edmunds, D. E., and Evans, W. D. Spectral Theory and Differential Operators. There they introduce the most common essential spectra and consider them in relative detail. The couterexample I mentioned eralier is also contained therein.
Best Answer
Check the following:
If a point $\lambda\in \sigma(A)$ is not an eigenvalue of $A$ then $\lambda$ is not isolated in $\sigma(A)$.
$\lambda$ is "isolated from" $\sigma(A)^2 \iff \pm\sqrt{\lambda}$ are both "isolated from" $\sigma(A)$. This tells you: $\lambda$ is not isolated from $\sigma(A)^2 \iff $ at least one of $\pm\sqrt\lambda$ is not isolated from $\sigma(A)$. Here "isolated from" just means isolated point except that the point is not necessarily in the set.
The eigenspace $E_\lambda(A^2) = E_{+\sqrt\lambda}(A) + E_{-\sqrt\lambda}(A)$, in particular if the lefthand side is an infinite dimensional space one of the spaces on the righthand side is infinite dimensional.
Now make use of $\sigma(A^2)=\sigma(A)^2$.
A point $\lambda$ is in $\sigma_{ess}(A^2)$ if and only if it is either not an eigenvalue of $A^2$ (in which case both of $\pm\sqrt\lambda$ are not an eigenvalue of $A$), it is not isolated from $\sigma(A^2)$ (in which case one of $\pm\sqrt\lambda$ is not isolated from $\sigma(A)$), or its eigenspace is infinite dimensional (in which case one of the eigenspaces of $\pm\sqrt\lambda$ must be infinite dimensional). This gives $\sigma_{ess}(A^2)\subseteq \sigma_{ess}(A)^2$.
A point $\lambda$ is in $\sigma_{ess}(A)^2$ if and only if either one of the roots $\pm\sqrt\lambda$ is not an eigenvalue (in which case that root is not isolated in $\sigma(A)$, hence its square $\lambda$ is not an isolated point of $\sigma(A^2)$), one of the roots is not isolated in $\sigma(A)$ (in which case the square also is not isolated), or the eigenspace to $A$ of one of the roots is infinite dimensional (in which case the eiegenspace to $A^2$ of $\lambda$ is also infinite dimensional). This gives $\sigma_{ess}(A)^2\subseteq \sigma_{ess}(A^2)$.