Is Not-STCON is NL-Complete? $STCON=\text{{(G,s,t)|G is a directed graph with a path from s to t}}$
$Co-STCON=\text{{(G,s,t)|G is a directed graph without a path from s to t}}$

I've tried the following proof:
Let $S\in NL$, and let $S^\complement \in coNL$. ($S^\complement$ complement is $S$)
Let $M_S$ be the NDTM  that solves $S$.
$x\in S \iff x\notin S^\complement$, so If I can decide membership in $S^\complement$, I can decide membership in $S$. Then I claim that $x\in S^\complement \iff(G_x,v_0,v_{acc}) \in Co-STCON$, where $G_x$ is the configuration graph of $M_S(x)$, $v_0$ is starting configuration, and $V_{acc}$ is a single accepting configuration. 

The problem I have with the proof above is that it has the form $x\in S \iff (G_x,v_0,v_{acc}) \notin Co-STCON$, so I feel like I'm missing something essential here. Is it correct?
Your proof doesn't work, since you're using an oracle reduction rather than a many-one reduction.

To show that a problem $A$ is NL-complete, you need to show that for every $S \in \mathsf{L}$ there is a logspace-computable function $f_S$ such that $x \in S$ iff $f_S(x) \in A$ (that's the definition); it's enough to show such a reduction for a single NL-complete problem $S$ (exercise). You show instead that $x \in S$ iff $f_S(x) \notin A$. 

Showing that nonSTCON is NL-complete is actually not so easy. Even showing that nonSTCON is in NL is not so easy; it is equivalent to showing that NL equals coNL. This is a non-trivial result due to Immerman and Szelepcsényi.