Your friends solution is correct. If $(X_t)_{t \geq 0}$ is a one-dimensional Itô process, then Itô's formula states $$ df(t,X_t)= \partial_x f(t,X_t) \, dX_t + \left(\frac{1}{2} \partial_x^2 f(t,X_t) \right) d\langle X \rangle_t + \partial_t f(t,X_t) \, dt. \tag{1}$$ Your friend used this identity for $f(t,x) := x e^{-ct}$.
Your attempt:
$\frac{\partial}{\partial W_t} \left( X_0 e^{ct} + \sigma e^{ct} \int_0^t e^{-cs} \, dW_s \right)$
It this how you are taught to write down Itô's formula? In my oppinion, that's not a good way to write it this way. The problem is that you cannot apply Itô's formula this way. Itô's formula gives you the differential for $f(t,W_t)$ for (nice) functions $f$. But here, you want to calculate the differential of the expression
$$\int_0^t e^{-cs} \, dW_s,$$
i.e. we need a function $f$ such that
$$f(t,W_t) \stackrel{!??!}{=} \int_0^t e^{-cs} \, dW_s.$$
... tell me: How do you choose $f$? Before you have not chosen such a function $f$, you cannot apply Itô's formula this way. What you are doing is treating it as a constant and that's simply not correct.
In order to solve this SDE (or check that the given process is a solution to the SDE) you really have to use Itô's formula for Itô proceses, i.e. $(1)$.
Remark The solution your friend suggested applies Itô's formula to the process
$$e^{-ct} X_t \tag{1}$$
and, at the first glance, it is not obvious how to come up with this particular process. The idea is the following: Instead of considering the SDE
$$dX_t = c X_t \, dt + \sigma \, dW_t$$
we consider the corresponding ODE
$$dx_t = cx_t \, dt$$
(i.e. we just we leave away the stochastic part). It is well-known that the unique solution to this ordinary differential equation is given by
$$x_t = C e^{ct}$$
where $C \in \mathbb{R}$. So far, $C$ is some "deterministic" constant. Now, however, we return to our stochastic setting and allow $C$ to depend on $\omega$ (this is the counterpart of the variation of constants-approach for SDEs). So, by the previous identity, our new auxilary process $C$ is given by
$$C_t = e^{-ct} X_t$$
... and this is exactly the process from $(1)$.
There are a lot of examples where this approach [i.e. first solve the corresponding ODE and then make a "stochastic" variation of constants] works, ee e.g. this question. However, I don't know any statements for which types of SDEs this approach works and for which it doesn't.
By Ito's lemma, we can compute that
$$dC_t = \alpha C_0 e^{\alpha W_t} dW_t + \frac{\alpha^2}{2}C_0 e^{\alpha W_t} dt$$
Since the covariation of a finite variation process and a semimartingale is $0$, we have that $$\langle X, C \rangle_t = \langle \int_0^\cdot \sigma X_s dW_s^* , \int_0^\cdot \alpha C_0 e^{\alpha W_s} dW_s \rangle = \int_0^t \alpha \sigma C_0 X_s e^{\alpha W_s} d\langle W_s^*, W_s \rangle = 0$$
where the final inequality follows since $W$ and $W^*$ are independent BMs.
Best Answer
First note that if $\mathbb{E}\left[X_t\mid Y=y\right]=g(y)$, for all $y$, then $\mathbb{E}\left[X_t\mid Y\right]=g(Y)$.
By definition, $$ X_t=\int_0^t b(X_s,Y)ds +\int_0^t \sigma(X_s,Y)dW_s $$ Denote by $\tilde{X}_t^y$ the solution to the SDE for a fixed $y$ (I included the dependency on y in the notation), then $$ \tilde{X}_t^y=\int_0^t b(\tilde{X}_s^y,y)ds +\int_0^t \sigma(\tilde{X}_s^y,y)dW_s $$
Thus, assuming $Y$ is independent on the Wiener process, $$ \mathbb{E}\left[ X_t \mid Y=y \right] = \mathbb{E}\left[ \int_0^t b(X_s,Y)ds +\int_0^t \sigma(X_s,Y)dW_s \mid Y=y \right] =\mathbb{E}\left[\tilde{X}_t^Y\mid Y=y \right] $$