The answer is *Yes*.

We are gonna use $\dim G = \dim F +2$ (the codimension of $F$ in $G$ is two) and $G\cap E=G'\cap E=F$.
You further assume that there exists $x\in\mathring F\subset \mathring G$. 
But we also have $x\in F\subset G'$, and the only way a face $G'$ can contain an interior point of a face $G$ is if $G\subseteq G'$.

The rest follows from dimension considerations:

\begin{align}
\dim G - 2 &= \dim F 
\\&= \dim(G'\cap E) 
\\&\ge n - ((n - \dim G') + \smash{\overbrace{(n - \dim E)}^{=2}})
\\&= \dim G' - 2,
\\[1em] &\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\implies \dim G'\le \dim G
\end{align}

where the inequality is the usual estimation for the dimension of the intersection of linear subspaces (technically, $G'$ is not a linear subspace; but locally at $x$, $G'$ is a "closed half-subspace" and thus intersects $E$ as a "full" subspace would).
Together with $G\subseteq G'$ we obtain $G=G'$.