Solution 1:

Part 1: Weak existence In order to construct and study weak solutions to SDEs it is often very useful to take advantage of the close connection between SDEs and martingale problems. For simplicity of notation I'll restrict myself to dimension $d=1$ but all the results below can be extended to higher dimensions.

Definition Define an operator $A$ by $$(Af)(t,x) := b(t,x) f'(x) + \frac{1}{2} \sigma^2 (t,x) f''(x), \qquad f \in C_b^2(\mathbb{R})$$ and denote by $X(t,\omega) := \omega(t)$ the canonical process on $C[0,\infty)$. We say that a probability measure $\mathbb{P}$ solves the $(A,\mu)$-martingale problem if $\mathbb{P}(X_0 \in \cdot) = \mu(\cdot)$ and $$M_t^f := f(X_t) - \int_0^t Af(s,X_s) \, ds$$ is a $\mathbb{P}$-martingale for all $f \in C_b^2(\mathbb{R})$.

Often the notation $\mathbb{P} = \mathbb{P}^{\mu}$ is used to indicate the initial distribution. It is not difficult to show (using Itô's formula) that any weak solution to the SDE

$$dY_t = b(t,Y_t) \, dt+ \sigma(t,Y_t) \, dB_t, \qquad X_0 \sim \mu, \tag{1}$$

gives rise to solution to the $(A,\mu)$-martingale problem ($\mathbb{P}$ is the distribution of $(Y_t)_{t \geq 0}$). It turns out that also the converse is true (see e.g. Ethier & Kurtz, Theorem 5.3.3).

Theorem Let $b$ and $\sigma$ be locally bounded and Borel measurable mappings, and let $\mu$ be a probability distribution. If there exists a solution to the $(A,\mu)$-martingale problem, then there exists a weak solution to $(1)$.

Now assume that for each $x \in \mathbb{R}$ the SDE (1) has a weak solution $(X_t^{(x)})_{t \geq 0}$ with initial distribution $\mu=\delta_x$. We denote by $\mathbb{P}^x$ the distribution of $(X_t^{(x)})_{t \geq 0}$ and define

$$\mathbb{P}^{\mu}(B) := \int_{\mathbb{R}} \mathbb{P}^x(B) \, \mu(dx)$$

for any measurable set $B \subseteq C[0,\infty)$. By the above theorem, it suffices to show that $\mathbb{P}^{\mu}$ is a solution to the $(A,\mu)$-martingale problem. Denote again by $(X_t)_{t \geq 0}$ the canonical process. Clearly,

$$\mathbb{P}^{\mu}(X_0 \in B) = \int_{\mathbb{R}} \underbrace{\mathbb{P}^x(X_0 \in B)}_{1_B(x)} \, \mu(dx) = \mu(B),$$

i.e. $X_0$ has initial distribution $\mu$. Moreover, we know that $(X_t^{(x)})_{t \geq 0}$ is a solution to the $(A,\delta_x)$-martingale problem and so

$$M_t := f(X_t)- \int_0^t Af(r,X_r) \, dr$$

is a $\mathbb{P}^x$-martingale for any $f \in C_b^2(\mathbb{R})$. This means that for any $s \leq t$ and any $F \in \mathcal{F}_s$ we have

$$\mathbb{E}^x(M_s 1_F) = \mathbb{E}^x(M_t 1_F). $$

Integrating both sides with respect to $\mu(dx)$ we get

$$\mathbb{E}^{\mu}(M_s 1_F) = \mathbb{E}^{\mu}(M_t 1_F),$$

and this shows that $(M_t)_{t \geq 0}$ is a $\mathbb{P}^{\mu}$-martingale. Applying the above theorem, we conclude that there exists a weak solution to $(1)$. Consequently, we have shown that it suffices to check the existence of weak solutions for initial distributions $\mu=\delta_x$. This result is, for instance, stated in Proposition 2 in the paper by Kallenberg "On the existence of universal functional solutions to classical SDEs"

Let me finally remark that also the existence of a unique weak solution can be characterized via martingale problems:

Let $\mu$ be a probability distribution. Uniqueness holds for the $(A,\mu)$-martingale problem if, and only if, the SDE (1) has a unique weak solution.

Part 2: Strong existence Let $(B_t)_{t \geq 0}$ be a Brownian motion and $\xi$ a random variable which is independent from $(B_t)_{t \geq 0}$. Suppose that the SDE

$$dY_t = b(t,Y_t) \, dt + \sigma(t,Y_t) \, dt, \qquad X_0=x$$

has a (strong) solution $(X_t^{(x)})_{t \geq 0}$ for any $x \in \mathbb{R}^d$ and that $(t,x,\omega) \mapsto X_t^{(x)}(\omega)$ is measurable. If we define

$$X_t := X_t^{(\xi(\omega))}(\omega).$$

then this gives a solution to the SDE with initial distribution $\xi$, see this question for details.