Consider the space $H = L^2([0,1],\mu)$, where $\mu$ is the Lebesgue measure. Define $T$ to be the multiplication with the identity function, i.e.
$$(Tf)(x) = x\cdot f(x).$$
Since the identity function is bounded, $T$ is bounded ($\lVert T\rVert \leqslant 1$), and since it is real-valued, $T$ is self-adjoint.
Clearly $T$ has no eigenvalues, since
$$(T - \lambda I)f = 0 \iff (x-\lambda)\cdot f(x) = 0 \quad \text{a.e.},$$
and since $x-\lambda$ is nonzero for all but at most one $x\in [0,1]$, it follows that $f(x) = 0$ for almost all $x$.
We have $\sigma(T) = [0,1]$, as is easily seen - for $\lambda \in \mathbb{C}\setminus [0,1]$, the function $x \mapsto \frac{1}{x-\lambda}$ is bounded on $[0,1]$.
The reason that $T$ has no eigenvalues is that for all $\lambda\in \mathbb{C}$ the set $\operatorname{id}_{[0,1]}^{-1}(\lambda)$ has measure $0$. If $m \colon [0,1] \to \mathbb{R}$ is a bounded measurable function such that $\mu(m^{-1}(\lambda)) > 0$ for some $\lambda$, then $\lambda$ is an eigenvalue of $T_m \colon f \mapsto m\cdot f$, any function $f\in H$ that vanishes outside $m^{-1}(\lambda)$ is an eigenfunction of $T_m$ then. Conversely, if $T_m$ has an eigenvalue $\lambda$ and $f \neq 0$ is an eigenfunction to that eigenvalue, then, since $(m(x) - \lambda)f(x) = 0$ almost everywhere, it follows that $f$ vanishes almost everywhere outside $m^{-1}(\lambda)$, and since $f\neq 0$, it further follows that $\mu(m^{-1}(\lambda)) > 0$.
If $T-\lambda I$ is injective, then $T-\lambda I$ will have an inverse on $\mathcal{R}(T-\lambda I)$, but that does not guarantee that $(T-\lambda I)^{-1} : \mathcal{R}(T-\lambda I)\subset X\rightarrow X$ is bounded. For example, consider $T : L^2[0,1]\rightarrow L^2[0,1]$ defined by
$$
Tf = \int_0^x f(t)dt.
$$
$T$ is bounded. Even though the inverse $T^{-1}g = g'$ is closed, it is defined only on functions $g \in L^2[0,1]$ that are
$\;\;\;$(i) absolutely continuous,
$\;\;\;$(ii) vanish at $0$, and
$\;\;\;$(iii) have a square-integrable derivative on $[0,1]$.
Furthermore $T^{-1}$ is not bounded on its domain; so it is not possible to extended $T^{-1}$ in such a way that it will be continuous. If the range of $T$ were all of $X$, so that the inverse of $T$ were defined everywhere on $L^2[0,1]$, then your argument would apply because $T$ would be defined on a Banach space and would have a closed graph. But that doesn't have to happen, even if $T^{-1}$ exists, as it does not happen in this case.
Best Answer
Without the assumption of normality, we can at least prove the following:
(This holds more generally in unital Banach algebras.) The second case is saying that $0$ is already in $\sigma(ST)$ and $\sigma(TS)$.
EDIT
This follows from the general statement:
Indeed, fix $\lambda\in\sigma(T)$. If $\lambda$ is an eigenvalue, we are done, so assume it is not, i.e., assume $\ker(T-\lambda)=\{0\}$. Since $T$ is normal, we have $\ker(T^*-\overline\lambda)=\{0\}$, and thus $$\overline{\operatorname{Range}(T-\lambda)}=\ker(T^*-\overline\lambda)^\perp=\mathcal{H}.$$ Hence $\operatorname{Range}(T-\lambda)$ is a proper dense subspace of $\mathcal{H}$. Thus $T-\lambda$ is not bounded below, and the result follows.