Let $u\in H^1(M)$. If $(u_n)$ is a sequence in $C_c^\infty(M)$ such that $u_n\to u$ in $H^1(M)$, then by definition $\widetilde{\mathrm{id}}(u)=\lim_{n\to\infty}u_n$ in $L^2(M)$, where existence and independence of this limit from the chosen sequence $(u_n)$ are standard exercises.
Therefore, to show that $\widetilde{\mathrm{id}}$ is injective, it suffices to show that whenever $(u_n)$ is an $H^1$-Cauchy sequence in $C_c^\infty(M)$ such that $u_n\to 0$ in $L^2(M)$, then $\|u_n\|_{H^1}\to 0$. In particular, $Du_n\to v$ for some $v\in L^2(M)$. Since $D$ is assumed to be symmetric on $C_c^\infty(M)$, it is closable (with closure $D^{\ast\ast}$). Hence $u_n\to 0$ and $Du_n\to v$ in $L^2(M)$ together imply $v=0$. Thus $\|u_n\|_{H^1}\to 0$.
Some remarks are in order: A quadratic form $q\colon D(q)\subset H\to H$ is called closable if the inclusion $D(q)\subset H$ extends to an injective linear map from the completion of $D(q)$ with respect to the norm $\|u\|_q=(\|u\|_H^2+q(u))^{1/2}$ into $H$. What I have shown above is essentially that if $q$ is of the form $q(u)=\|Su\|_H^2$ for some closable operator $S$ on $D(q)$, then $q$ is closable.
Note that I did not use that the manifold $M$ is closed nor that $D$ is elliptic, and instead of self-adjointness of $D$ it suffices to assume that $D$ is symmetric on $C_c^\infty(M)$, which is usually easier to verify than self-adjointness (although it is true that every symmetric elliptic differential operator on $C^\infty(M)$ is essentially self-adjoint if the manifold is closed).
