Let $$T$$ be a closed negative-definite symmetric densely defined unbounded operator on Hilbert space. I want to show that $$T$$ is self-adjoint if and only if $$T^*$$ has no positive eigenvalues.

For the direct way, because $$T$$ is closed symmetric densely defined and self-adjoint, we have that the spectrum of $$T$$ is a subset of $$\mathbb{R}$$. So if $$\lambda$$ is an eigenvalue and $$x$$ an eigenvector then $$0 > \langle Tx, x \rangle = \lambda \langle x,x \rangle.$$ Thus $$\lambda$$ must be negative.

I don't see what to do for the other way.

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 $$T^*=T$$.