Yes, it is. Let $C$ be a C*-algebra and let $A \subseteq C$ be an ideal which is intrinsically a von Neumann algebra. Then the positive part of the unit ball of $A$ has a least upper bound in $A$ which must be a projection. (Its norm cannot be greater than $1$, so if it is not a projection then its square root also belongs to the unit ball and is larger.) It follows that $C \cong A \oplus B$, so if $B$ is also intrinsically a von Neumann algebra then so is $C$.
I originally misread the question as asking whether $A$ and $C$ von Neumann implies $B$ von Neumann, which I think is actually a more interesting question, so I'll include that answer too:
Let $C$ be a von Neumann algebra and $A \subseteq C$ a C-ideal which is intrinsically a von Neumann algebra. We don't immediately know that $C/A$ is a von Neumann algebra because it isn't obvious that $A$ is weak closed in $C$. However, I claim that it is.
It will suffice to show that $A$ is normal, i.e., if $(x_\alpha)$ is a bounded increasing net of positive elements of $A$ then its least upper bound $x$ in $C$ belongs to $A$. Suppose not. The order structure is intrinsic to any C*-algebra, so $(x_\alpha)$ is also increasing in $A$ considered on its own as a von Neumann algebra. So $(x_\alpha)$ also has a least upper bound $y$ in $A$. We must show that $y = x$.
Suppose not. Since $y \geq x_\alpha$ for all $\alpha$, we have $y \geq x$. If they are not equal then $a = y - x$ is positive and nonzero. WLOG assume $y \leq 1$. Since $A$ is an ideal, we also have $b = a^{1/2}ya^{1/2} \in A$. Note that $b \leq a$, and $b$ can only equal zero if the range projection of $a^{1/2}$ (= the range projection of $a$) is in the kernel of $y$, which is impossible as $a \leq y$. Thus $b > 0$, and since it belongs to $A$, so does $y - b \geq x$. Since $x$ is the least upper bound of $(x_\alpha)$ in $C$, this shows that $y - b$ is an upper bound in $A$ and hence $y$ cannot be the least upper bound in $A$, a contradiction.