This is far from being true, indeed, every symmetric operator has only real eigenvalues:
If $\psi\in\ker(T-\lambda),\,\psi\neq 0$, then
$$
\lambda\|\psi\|^2=\langle T\psi,\psi\rangle=\langle \psi,T\psi\rangle=\bar\lambda\|\psi\|^2,
$$
hence $\lambda=\bar\lambda$.
Now every symmetric operator that is not self-adjoint yields a counterexample to your conjecture (if you want to be explicit, take $\Delta$ on $C_c^\infty(\mathbb{R}^n)$ as operator in $L^2(\mathbb{R}^n)$).
To get a criterion for self-adjointness, you have to replace the eigenvalues by the spectrum of the operator. Then the following characterization holds:
A symmetric operator $T$ is self-adjoint if and only if its spectrum is contained in $\mathbb{R}$.
Proof: It suffices to show that $D(T^\ast)\subset D(T)$. Let $z\in\mathbb{C}\setminus\mathbb{R}$. Since $\sigma(T)\subset\mathbb{R}$, the operators $T-z$ and $T-\bar z$ are invertible.
Let $\phi\in D(T^\ast)$ and $\psi:=(T-z)^{-1}(T^\ast -z)\phi\in D(T)$. Then we have $T\psi=T^\ast\psi$ and $(T^\ast-z)\psi=(T-z)\psi$.
It follows that
$$
(T^\ast-z)(\phi-\psi)=(T-z)\psi-(T-z)\psi=0,
$$
that is, $\phi-\psi\in N(T^\ast-z)=R(T-\bar z)^\perp=\{0\}$. Hence, $\phi=\psi\in D(T)$.
Remark: I used $R(A)$ and $N(A)$ to denote the range and kernel of $A$.
An elementary proof is possible, but it doesn't seem particularly enlightening. First some remarks: If $\lambda$ is in the residual spectrum then part of the definition is that $A-\lambda$ is injective, else you may find counterexamples (for example the $0$ operator). Secondly no self-adjoint operator may have a non-empty residual spectrum, hence any self-adjoint extension is a strict extension.
Finally the last preliminary remark, if $\mu$ is in the resolvent of $A$ and $A$ admits a strict self-adjoint extension $B$, then $\mu$ must be real. For $A-\mu: D(A)\to H$ must be bijective, but if $\mu$ is not real then $B-\mu$ is invertible (since $B$ self-adjoint) and hence $B-\mu : D(B)\to H$ must be bijective, even though it already admits a surjective restriction to a proper subset, contradicting injectivitiy.
Now let $\lambda\in\sigma_{res}(A)$, $\mu\in\rho(A)$ and suppose $B$ is a self-adjoint extension of $A$. Since $\overline{(A-\lambda I)H}\neq H$ is closed, there is a non-zero $z\in H$ so that $z$ is orthogonal to the range of $A-\lambda$, ie
$$\langle z , (A-\lambda )y \rangle =0 \quad \text{ for all $y\in D(A)$}.$$
First we remark that $z\notin D(A)$. If $z\in D(A)$ you get that $\langle (A-\overline\lambda )z, y\rangle = \langle z, (A-\lambda)y\rangle= 0$ for all $y\in D(A)$, which is dense in $H$, so $(A-\overline\lambda)z=0$, meaning that $\overline\lambda$ is an eigenvalue of $A$ and $z$ is an eigenvector. Since $A$ is symmetric you immediately find that $\lambda$ must be real. So you get $(A-\lambda)z=0$, contradicting injectivity of $A-\lambda$, finally giving $z\notin D(A)$.
Now we consider two cases: either $z\in D(B)$ or $z\notin D(B)$. Both will yield a contradiction.
If $z\in D(B)$ then by symmetry of $B$ you have $\langle (B-\lambda) z , y\rangle = \langle z , (A-\lambda)y \rangle = 0$ for all $y\in D(A)$. Hence $Bz = \lambda z$. Now let $w\in D(A)$ with $(A-\mu)w= (\lambda-\mu)z$. Then
$$\langle z - w , (A-\mu) y \rangle = \langle (\lambda - \mu)z - (\lambda-\mu )z , y\rangle = 0 $$
for all $y\in D(A)$. Since $(A-\mu)D(A)=H$ you then get $z=w$, hence $z\in D(A)$ must already hold, which we have already seen is not allowed.
If $z\notin D(B)$ then by self-adjointness of $B$ you get $z\notin D(B^*)$, hence there must be some sequence of norm one vectors $\xi_n\in D(B)$ with $\langle z, B\xi_n\rangle$ being unbounded. Now let $w_n$ be such that $(A-\mu)w_n = (B-\mu)\xi_n$ and $w$ such that $(A-\mu)w=z$. First note that:
$$\langle w, (B-\mu)\xi_n\rangle = \langle z, \xi_n\rangle$$
hence $\langle w, (B-\mu)\xi_n\rangle$ is bounded. On the other hand:
$$\langle z, (B-\mu)\xi_n\rangle = \langle z, (A-\lambda)w_n+(\lambda-\mu)w_n\rangle=\langle z, (\lambda-\mu)w_n\rangle =\langle ( A-\mu) w, (\lambda-\mu)w_n\rangle \\ = \langle w,(\lambda -\mu)(A-\mu)w_n\rangle = (\lambda-\mu) \langle w, (B-\mu)\xi_n\rangle$$
whence $\langle w, (B-\mu)\xi_n\rangle$ must be unbounded. Contradiction
Best Answer
Suppose $T$ is symmetric, and has a unique selfadjoint extension. (I'll assume that $T$ is densely-defined, though you can probably prove that it must be the case.)
Let $\overline{T}$ be the closure of $T$, which exists because $T$ is symmetric. $\overline{T}$ must also be symmetric, and because $\overline{T}^*=T^*$, then $$ \mathcal{D}(T^*)=\mathcal{D}(\overline{T})\oplus\mathcal{N}(T^*-iI)\oplus\mathcal{N}(T^*+iI), $$ where the decomposition is orthogonal with respect to the graph inner product $$ (x,y)_{T^*} = (x,y)_H+(T^*x,T^*y)_{H}. $$ If $S$ is a selfadjoint extension of $T$, then $S$ is also a selfadjoint extension of $\overline{T}$ because the graph of $T$ is contained in the graph of $S$ and the graph of $S$ is closed. $\overline{T}$ has a selfadjoint extension iff $\mathcal{N}(T^*-iI)$ is unitarily equivalent to $\mathcal{N}(T^*+iI)$, and the possible selfadjoint extensions are in one-to-one correspondence with the unitary maps $U: \mathcal{N}(T^*-iI)\rightarrow\mathcal{N}(T^*+iI)$. However, because $\overline{T}$ has only one selfadjoint extension, then $\mathcal{N}(T^*-iI)=\{0\}=\mathcal{N}(T^*+iI)$ follows. Hence, $\overline{T}=T^*$, which also gives $\overline{T}=\overline{T}^*$. So $\overline{T}$ must be selfadjoint.