Question
Is it true in general that if you have a multidimensional process: $$ \begin{align} \mathrm{d}X_t &= \mu(X_t,Y_t)\,\mathrm{d}t + \sigma(X_t,Y_t)\,\mathrm{d}V_t \\ \mathrm{d}Y_t &= \nu(Y_t)\,\mathrm{d}t + \tau(Y_t)\,\mathrm{d}W_t \end{align} $$ then $X_t$ follows the SDE: $$ \mathrm{d}X_t = \mathbb{E}[\mu(X_t,Y_t)|X_t]\,\mathrm{d}t + \sqrt{\mathbb{E}[\sigma^2(X_t,Y_t)|X_t]}\,\mathrm{d}V_t $$ If not, is there well-defined circumstances under which that is true?
Example
I can see that it works in the following case. Consider a 2-dimensional Ornstein-Uhlenbeck process: $$ \begin{align} \mathrm{d}X_t &= -(a X_t + b Y_t)\,\mathrm{d}t + \sigma\,\mathrm{d}V_t \\ \mathrm{d}Y_t &= -c Y_t\,\mathrm{d}t + \tau\,\mathrm{d}W_t \end{align} $$ with $a,b,c,\sigma,\tau > 0$. Define: $$ A=\begin{bmatrix}a & b \\ 0 & c\end{bmatrix} \qquad B=\begin{bmatrix}\sigma & 0 \\ 0 & \tau\end{bmatrix} $$ so that we have, using vector notation, $\mathrm{d}Z_t = -A Z_t\,\mathrm{d}t + B\,\mathrm{d}W_t$ where $Z_t = (X_t,Y_t)$. Since $A$ has two positive eigenvalues ($a$ and $c$), the process admits a stationary multivariate normal distribution with mean $0$ and variance matrix: $$ \Sigma = \left[ \begin{array}{cc} \frac{c (a+c) \sigma ^2+b^2 \tau ^2}{2 a c (a+c)} & -\frac{b \tau ^2}{2 c (a+c)} \\ -\frac{b \tau ^2}{2 c (a+c)} & \frac{\tau ^2}{2 c} \\ \end{array} \right] $$ (see Gardiner (2008), Stochastic Methods, section 4.5.6 for details).
On the one hand, the marginal stationary distribution of $X_t$ is normal with mean zero and variance: $$\frac{c \sigma ^2 (a+c)+b^2 \tau ^2}{2 a c (a+c)}$$
On the other hand, the conditional expectation of $Y_t|X_t$ is: $$ \mathbb{E}[Y_t|X_t]=-X_t\frac{a b \tau ^2}{c \sigma ^2 (a+c)+b^2 \tau ^2} $$ and therefore: $$ \mathbb{E}[aX_t+bY_t|X_t]=X_t\frac{a c \sigma ^2 (a+c)}{c \sigma ^2 (a+c)+b^2 \tau ^2} $$ So the 1-dimensional Ornstein-Uhlenbeck process: $$ \mathrm{d} U_t = \mathbb{E}[-(aX_t+bY_t)|X_t] \,\mathrm{d}t + \sigma\,\mathrm{d}V_t = -X_t\frac{a c \sigma ^2 (a+c)}{c \sigma ^2 (a+c)+b^2 \tau ^2}\,\mathrm{d}t + \sigma\,\mathrm{d}V_t $$ has a normal stationary distribution with variance $\frac{c \sigma ^2 (a+c)+b^2 \tau ^2}{2 a c (a+c)}$, the same as $X_t$ above.
Attempt for the general case
To try to prove the general case, I started from the Fokker–Planck equation: $$ \frac{\partial}{\partial t}f_t(x,y) = -\frac{\partial}{\partial x} [\mu(x,y)f_t(x,y)] -\frac{\partial}{\partial y} [\nu(y)f_t(x,y)] + \frac{1}{2} \frac{\partial^2}{\partial x^2} [\sigma^2(x,y)f_t(x,y)] + \frac{1}{2} \frac{\partial^2}{\partial y^2} [\tau^2(y)f_t(x,y)] $$ Then I integrate each term with respect to $y$. For the terms in $x$, I get the expected result quite easily: $$ \begin{align} \int \frac{\partial}{\partial t}f_t(x,y)\,\mathrm{d}y &= \frac{\partial}{\partial t}\int f_t(x,y)\,\mathrm{d}y = \frac{\partial}{\partial t} f_t(x) \\ \int \frac{\partial}{\partial x} [\mu(x,y)f_t(x,y)]\,\mathrm{d}y &= \frac{\partial}{\partial x} \int [\mu(x,y)f_t(x,y)]\,\mathrm{d}y = \frac{\partial}{\partial x} \{\mathbb{E}[\mu(X_t,Y_t)|X_t=x]f_t(x)\} \\ \int \frac{\partial^2}{\partial x^2} [\sigma^2(x,y)f_t(x,y)]\,\mathrm{d}y &= \frac{\partial^2}{\partial x^2} \int[\sigma^2(x,y)f_t(x,y)]\,\mathrm{d}y = \frac{\partial^2}{\partial x^2} \{\mathbb{E}[\sigma^2(X_t,Y_t)|X_t=x]f_t(x)\} \end{align} $$ But then I have to get rid of the terms in $y$, and I'm a bit stuck. I guess that I can get rid of $\int \frac{\partial}{\partial y} [\nu(y)f_t(x,y)]\,\mathrm{d}y$ using $\nu(y) = \mathcal{O}(y)$ and $f(x,y) = \mathcal{o}(y)$. But I'm less sure of what to do for $\int \frac{\partial^2}{\partial y^2} [\tau^2(y)f_t(x,y)]\,\mathrm{d}y$. Using $\sigma(y)=\mathcal{O}(y)$ will work for when $f_t$ is normal as in the Ornstein-Uhlenbeck case. It can also work when $f(y) \sim y^{-\beta}$, $\beta > 1$. But you need the derivative of $f_t$ to converge towards zero, and you can obviously construct densities for which that is not the case. At the same time, those densities would have irregularly varying tails, and I'm not sure they can be the solution of a SDE. I tried substituting using the Fokker–Planck equation for $Y_t$ alone, but then there's a term $f_t(x|y)$ that I can't get rid of.