I am trying to prove: Given two continuous functions $f,b$, and $X$ a stochastic process on $(\Omega,\mathcal{F},P)$, satisfying $X_t=\int_0^tb(X_s)ds+\int_0^tf(X_s)dW_s$, if $\hat X$ is a stochastic process on $(\hat\Omega,\mathcal{\hat F},\hat P)$, which has the same distribution as $X$, then we have $\hat X_t=\int_0^tb(\hat X_s)ds+\int_0^tf(\hat X_s)d\hat W_s$.
But I don't know if the left parts have the same distribution, or each the integral has the same distribution.
I tried to prove the stochastic integrals are distributed the same by assuming $X$ and $\hat X$ are simple processes but failed.
Edit: See Kurt's comment below for a way to take this construction and account for the continuity constraint.
You cannot prove it because it is not true. Take $\hat{X}$ and $W$ two independent Brownian motions and define $X_t = \int_0^t \mathrm{sgn}(W_s) dW_s$. Seeing that $X_t$ is a continuous martingale with linear quadratic variation, Levy's characterisation theorem tells us it is a Brownian motion, so it agrees in distribution with $\hat X$. Define $$\hat{Y} = \int_0^t \mathrm{sgn}(\hat{X}_s)dW_s$$ Then, $\hat{Y} \neq \hat{X}$ because $\langle \hat{X}, Y \rangle_t = 0 \neq t = \langle X \rangle_t$.