On a finite interval $[0,T]$, consider a standard Brownian motion $W$ and a stochastic differential equation $dX = \mu(X) \, dt + \sigma(X) \, dW$. I would like to know when one can define a mapping $\Phi: \mathbb{R} \times C([0,T], \mathbb{R}) \to C([0,T], \mathbb{R})$ that, for an initial condition $x_0 \in \mathbb{R}$ and a Brownian path $w \in C([0,T], \mathbb{R})$, produces a path $\Phi(x_0, w) \in C([0,T]$ such that the random variable $\Phi(x_0, W)$ has the same law as the SDE with initial condition $x_0$.
In the case when $\sigma$ is a constant and $\mu$ is globally Lipshitz, such a function $\Phi$ exists and is even Lipshitz,
$$ \|\Phi(x_0^{(2)}, w^{(2)}) - \Phi(x_0^{(1)}, w^{(1)})\|_{[0,T], \infty} \leq \textrm{Cst} \times \left\{ |x_0^{(1)} - x_0^{(2)}| + \|w^{(1)} - w^{(2)}\|_{[0,T], \infty} \right\}. $$
For seeing this, it suffices to substract the quantities $$ x^{i}_0 + \int_{0}^t \mu(x^{(i)}_s) \, ds + \sigma \, w^{(i)}_t $$ for $i=1,2$ and invoke Gronwall's lemma. Can this be generalized to the case when the function $\sigma(\cdot)$ is not constant?
When discretizing SDEs (eg. Euler-Maruyama and variants), everything can be done in a path-wise manner. Is it true that the same remains true in the continuous limit?
The existence of the functional $\Phi$ is often taken as a definition of a strong solution. Your question boils down to "when does a strong solution exist?". One thing to look up is the classic result of Yamada-Watanabe by which weak existence and pathwise uniqueness imply strong existence and the uniqueness in probablility. See for example the book Karatzas & Shreve, Brownian Motion and Stochastic Calculus or the papers [1] and [2].