I am writing down a very similar answer to this one, that I wrote for the similar question *SDE with non-degenerate diffusion visits every point*. I am not sure how closely it answers your question, since it is not a reference for the exact result you quote, but it does follow as a corollary.

The support of the whole trajectory $X$ in $\mathcal C([0,\infty),\mathbb R^n)$ is described by the so-called Stroock-Varadhan support theorem. In *On the support of diffusion processes with applications to the strong maximum principle*, their Theorem 3.1 ensures that the (unique) solution to the martingale problem associated with $u_t+\frac12\Delta_K$ has full support provided $u:[0,\infty)\times\mathbb R^d\to\mathbb R^d$ is bounded measurable.

A strong solution to the equation $\mathrm dX_t=u_t(X_t)\mathrm dt + \mathrm dB_t$ will always be a solution to the martingale problem, so this answers your question in the case where (1) you have conditions guaranteeing existence of a strong solution and (2) $u$ is bounded measurable (you can also define the solution as a solution to the martingale problem, in which case (2) is sufficient for existence and uniqueness).

A natural case that doesn't quite fit the above is when $u$ is unbounded but you have a unique strong and weak solution up to some explosion time using some other argument $\tau$. We can still rely on this theorem in more general situations, and to illustrate my point I will consider the classical case where $u$ is continuous in $(t,x)$, locally Lipschitz in $x$ (for instance $u$ is $\mathcal C^1$ in $(t,x)$). Then one can define the solution $X^R>0$ to the equation
$$ \mathrm dX^R_t = \big(0\vee(2R-|X^R_t|)\wedge1\big)u_t(X^R_t)\mathrm dt + \mathrm dB_t, $$
which is defined for all times and coincides with $X$ until one of the two exists the ball of radius $R$. This means that
$$\mathbb P(X_t\in U)\geq\mathbb P(X_t\in U\text{ and }\forall s\leq t,|X_s|<R)=\mathbb P(X^R_t\in U\text{ and }\forall s\leq t,|X^R_s|<R)$$
for $R$ large enough. By the theorem above, we know that this last probability must be positive, and so is the first one.

A similar argument shows that your condition $|u_s(x)|\leq C_t(1+|x|)$ is enough, provided $u$ is measurable, and your notion of solution coincides with the unique martingale problem solution.

I should add that these results rely on applying a Girsanov argument, making a change of probability with density (if I am not mistaken with my use of $K$ and $K^{-1}$)
$$\exp\left(\int_0^tu_s(X_s)K^{-1}\mathrm dB_s-\frac12\int_0^tu_s\mathrm ds\right).$$