Let $H$ be the underlying Hilbert space, $I$ the identity operator on $H$, $V=\operatorname{Range}(T-I)$ and $\overline{V}$ the closure of $V$. It is a standard result of operators on Hilbert space that $V^\perp = \operatorname{Ker}((T-I)^*) = \operatorname{Ker}(T^*-I)$. Since $T^*$ is assumed to have no positive eigenvalues, $V^\perp = \operatorname{Ker}(T^*-I)=0$. Hence $V$ is dense in $H$.
For any $u\in\mathcal{D}(T)$,
$$\|(T-I)u\|^2 = \langle Tu-u,Tu-u\rangle = \|Tu\|^2+\|u\|^2 - \langle Tu,u\rangle - \langle u,Tu\rangle = \|Tu\|^2+\|u\|^2 - 2\langle u,Tu\rangle$$
where $\langle Tu,u\rangle = \langle u,Tu\rangle$ by the symmetry of $T$. Since $\langle u,Tu\rangle \le 0$ by assumption, we conclude
$$\|(T-I)u\| \ge \|u\|$$
It follows that not only is $(T-I)$ one-to-one, but it is bounded below. So considering the inverse $(T-I)^{-1}$ defined on $V$, it is a bounded linear operator. If $z\in\overline{V}$ there exists a sequence $z_n\in V$ such that $z_n\rightarrow z$. Then $z_n$ is Cauchy. Letting $y_n = (T-I)^{-1}(z_n)$, we have that $y_n$ is Cauchy because $(T-I)^{-1}$ is bounded. Let $y$ be the limit of $y_n$. then $y_n\rightarrow y$ and $z_n=(T-I)y_n \rightarrow z$. Since $T$ is closed, so is $(T-I)$. Therefore $z\in V$ and $z=(T-I)y$. Hence $\overline{V} = V$. We already showed that $V$ is dense, so $V=H$.
Now, $T$ is symmetric, so $T^*$ is an extension of $T$, so $(T^*-I) = (T-I)^*$ is an extension of $(T-I)$. But $V=\operatorname{Range}(T-I)$ is already all $H$, so either $(T^*-I)$ isn't one-to-one, or $(T^*-I) = (T-I)$. We know however from assumption that $\operatorname{Ker}(T^*-I) = 0$. So $(T^*-I)$ is one-to-one, and it follows that $(T^*-I) = (T-I)$. From this it follows that $T^*=T$.
