There is some redundant work, but not as much as you were hoping. Recall the $LDL^T$ factorization
$$C = \begin{pmatrix} A & B^T \\ B & D \end{pmatrix}
=\begin{pmatrix} 1 & 0 \\ B A^{-1} & 1 \end{pmatrix}
\begin{pmatrix} A & 0 \\ 0 & S \end{pmatrix}
\begin{pmatrix} 1 & A^{-1} B^T \\ 0 & 1 \end{pmatrix}$$
where $S = D - B A^{-1} B^T$ is the Schur complement. Clearly you need $A^{-1}$ and $S^{-1}$ to solve using this factorization, with $D^{-1}$ providing no value.
Substituting $A = L_A L_A^T$ and $S = L_S L_S^T$ into the factorization above, we obtain the Cholesky factorization [1]:
$$\begin{aligned}
C & =\begin{pmatrix}
A & B^{T}\\
B & D
\end{pmatrix}\\
& =\begin{pmatrix}
\mathbb{1} & 0\\
B\left( L_{A} L_{A}^{T}\right)^{-1} & \mathbb{1}
\end{pmatrix}\begin{pmatrix}
L_{A} L_{A}^{T} & 0\\
0 & L_{S} L_{S}^{T}
\end{pmatrix}\begin{pmatrix}
\mathbb{1} & \left( L_{A} L_{A}^{T}\right)^{-1} B^{T}\\
0 & \mathbb{1}
\end{pmatrix}\\
& =\left(\begin{pmatrix}
\mathbb{1} & 0\\
B\left( L_{A} L_{A}^{T}\right)^{-1} & \mathbb{1}
\end{pmatrix}\begin{pmatrix}
L_{A} & 0\\
0 & L_{S}
\end{pmatrix}\right)\left(\begin{pmatrix}
L_{A}^{T} & 0\\
0 & L_{S}^{T}
\end{pmatrix}\begin{pmatrix}
\mathbb{1} & \left( L_{A} L_{A}^{T}\right)^{-1} B^{T}\\
0 & \mathbb{1}
\end{pmatrix}\right)\\
& =\begin{pmatrix}
L_{A} & 0\\
BL_{A}^{-T} & L_{S}
\end{pmatrix}\begin{pmatrix}
L_{A}^{T} & L_{A}^{-1} B^{T}\\
0 & L_{S}^{T}
\end{pmatrix} =L_{C} L_{C}^{T} .
\end{aligned}
$$
Thus you can simply extract the factors of $A$ from the factors of $C$, but will have to factor $D$ separately. If the block sizes are equal, $C$ takes eight times more work to factor, so this is little savings.
If you have a sequence of problems and can partition your system so that $A$ does not change between subsequent solves, its factorization could be reused, but $S$ will change and have to be refactored. Depending on the decomposition and conditioning, iterative methods can be useful to reuse a partial factorization.
[1] Note that this "block Cholesky" factorization is also valid with $L_A$ replaced by a $A^{1/2}$ or another non-triangular factorization of $A$.
