The answer is *Yes*.

Assume $\dim G = \dim F +2$ (the codimension of $F$ in $G$ is two) and $G\cap E=G'\cap E=F$.

You also assumed 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 $G\subseteq G'$.

For convenience we translate the setup so that $x=0$.
Then $E$ is still a *linear* subspace and we choose a basis $e_1,...,e_{n-2}\in\Bbb R^n$ of $E$. 
Since $F\subset E$ we can assume that $e_1,...,e_{\dim F}\in F\subseteq G\subseteq G'$. 
But we also need to assume that $e_{\dim F+1},...,e_{n-2}\perp G'$, as otherwise $G'$ would have intersections with $E$ other than $F$.
From this we can estimate the dimension of $G'$:

\begin{align}
\dim G' &\le n-((n-2)-(\dim F+1)+1) = \dim F+2 = \dim G
\end{align}

Together with $G\subseteq G'$ we obtain $G=G'$.