The answer is *Yes*.

By your assumption there exists $x\in\mathring F\subset \mathring G$.
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 $G\subseteq G'$.

Now, consider a small neighborhood of $x$. In this neighborhood, $F$, $G$ and $E$ look like linear subspaces, while $G'$ looks like a cone that contains the subspace $G$. We will keep this local perspective for the rest of this proof (we can assume that $x$ is the origin of our local linear space).

Since all subspaces contain $F$, we can factor by $F$. We find 

$$G/F\cap E/F = (G\cap E)/F=F/F = \{0\}.$$

Moreover, since $\DeclareMathOperator{\codim}{codim}\codim_G F=2$, we have $\dim G/F=2$; and since $\codim_{\Bbb R^n} E=2$, we have $\codim_{\Bbb R^n/F} E/F=2$. Therefore,

$$G/F \oplus E/F = \Bbb R^n/F.$$

We now show that also $G'\subseteq G$: fix $y\in G'/F$.
Then there is a unique decomposition $y=y'+y''$ with $y'\in G/F$ and $y''\in E/F$. Since $y'\in G/F$ and $G/F$ is a linear space, we have $-y'\in G/F\subset G'/F$.
Since $G'/F$ is a cone, we have $y+(-y')=y''\in G'/F$. Thus

$$y''\in E/F\cap G'/F = (E\cap G')/F=F/F=\{0\}$$

and we conclude $y=y'\in G/F$. Thus $G=G$'.