There are two questions. First, there is a purely mathematical question about the possibility of decomposing the GLS estimator into the OLS estimator plus a correction factor. Second, there is a question about what it means when OLS and GLS are the same. (I will use ' rather than T throughout to mean transpose).
Also, I would appreciate knowing about any errors you find in the arguments.
Question 1
Ordinary Least Squares (OLS) solves the following problem:
\begin{align}
min_x\;\left(y-Hx\right)'\left(y-Hx\right)
\end{align}
leading to the solution:
\begin{align}
\hat{x}_{OLS}=\left(H'H\right)^{-1}H'y
\end{align}
Generalized Least Squares (GLS) solves the following problem:
\begin{align}
min_x\;\left(y-Hx\right)'C^{-1}\left(y-Hx\right)
\end{align}
leading to the solution:
\begin{align}
\hat{x}_{OLS}=\left(H'C^{-1}H\right)^{-1}H'C^{-1}y
\end{align}
Now, make the substitution $C^{-1}=X+I$ in the GLS problem:
\begin{align}
min_x\;&\left(y-Hx\right)'\left(X+I\right)\left(y-Hx\right)\\~\\
min_x\;&\left(y-Hx\right)'X\left(y-Hx\right) + \left(y-Hx\right)'\left(y-Hx\right)\\
\end{align}
The solution is still characterized by first order conditions since we are assuming that $C$ and therefore $C^{-1}$ are positive definite:
\begin{align}
0=&2\left(H'XH\hat{x}_{GLS}-H'Xy\right) +2\left(H'H\hat{x}_{GLS}-H'y\right)\\
\hat{x}_{GLS}=&\left(H'H\right)^{-1}H'y+\left(H'H\right)^{-1}H'Xy
-\left(H'H\right)^{-1}H'XH\hat{x}_{GLS}\\
\hat{x}_{GLS}=& \hat{x}_{OLS} + \left(H'H\right)^{-1}H'Xy
-\left(H'H\right)^{-1}H'XH\hat{x}_{GLS}\\
\end{align}
I can see two ways to give you what you asked for in the question from here. First, we have a formula for the $\hat{x}_{GLS}$ on the right-hand-side of the last expression, namely $\left(H'C^{-1}H\right)^{-1}H'C^{-1}y$. Thus, the above expression is a closed form solution for the GLS estimator, decomposed into an OLS part and a bunch of other stuff. The other stuff, obviously, goes away if $H'X=0$. To be clear, one possible answer to your first question is this:
\begin{align}
\hat{x}_{GLS}=& \hat{x}_{OLS} + \left(H'H\right)^{-1}H'Xy
-\left(H'H\right)^{-1}H'XH\left(H'C^{-1}H\right)^{-1}H'C^{-1}y\\
\hat{x}_{GLS}=& \hat{x}_{OLS} + \left(H'H\right)^{-1}H'X \left(I
-H\left(H'C^{-1}H\right)^{-1}H'C^{-1}\right)y
\end{align}
I can't say I get much out of this. That awful mess near the end multiplying $y$ is a projection matrix, but onto what?
Another way you could proceed is to go up to the line right before I stopped to note there are two ways to proceed and to continue thus:
\begin{align}
\left(I+\left(H'H\right)^{-1}H'XH\right)\hat{x}_{GLS}=& \hat{x}_{OLS} + \left(H'H\right)^{-1}H'Xy\\
\hat{x}_{GLS}=& \left(I+\left(H'H\right)^{-1}H'XH\right)^{-1}\left(\hat{x}_{OLS} + \left(H'H\right)^{-1}H'Xy\right)
\end{align}
Again, GLS is decomposed into an OLS part and another part. The other part goes away if $H'X=0$. I still don't get much out of this. What this one says is that GLS is the weighted average of OLS and a linear regression of $Xy$ on $H$. I guess you could think of $Xy$ as $y$ suitably normalized--that is after having had the "bad" part of the variance $C$ divided out of it.
I should be careful and verify that the matrix I inverted in the last step is actually invertible:
\begin{align}
\left(I+\left(H'H\right)^{-1}H'XH\right) &= \left(H'H\right)^{-1}\left(H'H+H'XH\right)\\
&= \left(H'H\right)^{-1}H'\left(I+X\right)H\\
&= \left(H'H\right)^{-1}H'C^{-1}H
\end{align}
Question 2
The question here is when are GLS and OLS the same, and what intuition can we form about the conditions under which this is true? I will only provide an answer here for a special case on the structure of $C$. The requirement is:
\begin{align}
\left(H'C^{-1}H\right)^{-1}H'C^{-1}Y = \left( H'H\right)^{-1}H'Y
\end{align}
To form our intuitions, let's assume that $C$ is diagonal, let's define $\overline{c}$ by $\frac{1}{\overline{c}}=\frac{1}{K}\sum \frac{1}{C_{ii}}$, and let's write:
\begin{align}
\left(H'C^{-1}H\right)^{-1}H'C^{-1}Y &=
\left(H'\overline{c}C^{-1}H\right)^{-1}H'\overline{c}C^{-1}Y\\
&=\left( H'H\right)^{-1}H'Y
\end{align}
One way for this equation to hold is for it to hold for each of the two factors in the equation:
\begin{alignat}{3}
\left(H'\overline{c}C^{-1}H\right)^{-1}
&=\left( H'H\right)^{-1} & \iff& & H'\left(\overline{c}C^{-1}-I\right)H&=0\\
H'\overline{c}C^{-1}Y&=H'Y & \iff& & H'\left(\overline{c}C^{-1}-I\right)Y&=0
\end{alignat}
Remembering that $C$, $C^{-1}$, and $I$ are all diagonal and denoting by $H_i$ the $i$th row of $H$:
\begin{alignat}{3}
H'\left(\overline{c}C^{-1}-I\right)H&=0 & \iff&
& \frac{1}{K} \sum_{i=1}^K H_iH_i'\left( \frac{\overline{c}}{C_{ii}}-1\right)=0\\~\\
H'\left(\overline{c}C^{-1}-I\right)Y&=0 & \iff&
& \frac{1}{K} \sum_{i=1}^K H_iY_i\left( \frac{\overline{c}}{C_{ii}}-1\right)=0
\end{alignat}
What are those things on the right-hand-side of the double-headed arrows? They are a kind of sample covariance. To see this, notice that the mean of $\frac{\overline{c}}{C_{ii}}$ is 1, by the construction of $\overline{c}$. Finally, we are ready to say something intuitive. In this special case, OLS and GLS are the same if the inverse of the variance (across observations) is uncorrelated with products of the right-hand-side variables with each other and products of the right-hand-side variables with the left-hand-side variable. This is a very intuitive result.
In estimating the linear model, we only use the products of the RHS variables with each other and with the LHS variable, $(H'H)^{-1}H'y$. In GLS, we weight these products by the inverse of the variance of the errors. When does that re-weighting do nothing, on average? Why, when the weights are uncorrelated with the thing they are re-weighting! Yes? When is a weighted average the same as a simple average? When the weights are uncorrelated with the things you are averaging.
This insight, by the way, if I am remembering correctly, is due to White(1980) and perhaps Huber(1967) before him---I don't recall exactly.
Best Answer
Pretending to know the true variance is always a funny thing to see. I would never trust that sort of an assumption, unless it comes from a carefully designed experiment where an artificial correlation structure was imposed (although frankly I cannot really imagine a practical set up that would lead to this).
From the GEE perspective, you should be able to get more efficient estimates when your assumption about $\bf V$ is correct, as compared to OLS, but you would still want to use the sandwich variance estimator in the (highly unlikely) case that you are mistaken about the covariances. Usually, robustness to model assumptions is considered a greater issue than efficiency, unless you have really tiny sample sizes, and any 20% efficiency gain is a big deal. So I would run this with as the GLS (or feasible GLS, if you only know the structure of $\bf V$, but not the specific parameter values), but still correct for clustering using the sandwich variance estimator.