Let the SDE $$dX_t=\mu(X_t)dt+\sigma (X_t)dB_t,\quad X_0=x_0,$$
for $\mu$ and $\sigma $ nice enough, why do we always consider $p(x,t|x_0)$ for the density function and never $p(x,t)$ ? Why is it always a conditional density ? The density $p(x,t)$ doesn't makes sense or what ?
By the way, since $\mathbb P(X_0=x_0)=1$, don't we have that $$p(x,t)dx=\mathbb P(X_t \in dx)=\mathbb P(X_t\in dx, X_0=x_0)=\mathbb P(X_t\in dx|X_0=x_0)P(X_0=x_0)=p(x,t|x_0)dx \ \ ?$$
The stochastic differential equation $$ \mathrm d X_t = \mu(X_t) \mathrm d t + \sigma(X_t) \mathrm d B_t,$$ is just an informal way to represent the integral equation $$ X_{t+h} - X_t = \int_t^{t+h} \mu(X_s) \mathrm d s + \int_t^{t+h}\sigma(X_s) \mathrm d B_s,$$ where the first integral is a Lebesgue integral and the second is an Ito integral.
When we say that $X_t$ solves the SDE, what we really mean is that $X_t$ solves the integral equation above. Now you can see that the right-hand side is $X_\tau$-measurable, so it is a random variable that depends on the value of $X_\tau$ (that is, when we start looking at the integral equation).
One difference between SDEs and ODEs is that SDEs have two types of solutions.
The stochastic process $X_t$ with the filtration $\mathcal F_t$ is called a strong solution with initial condition $x_0$ if $$ X_{t} - X_0 = \int_0^{t} \mu(X_s) \mathrm d s + \int_0^{t} \sigma(X_s) \mathrm d B_s \quad X_0 = x_0,$$ holds almost surely for all $t\geq 0$.
The stochastic process $X_t$ is a weak solution with initial distribution $\nu$ if there is a Brownian motion $B_t$ such that $\mathcal F_t$ is an admissible filtration, $\mathbb P(X_0 \in \cdot) = \nu(\cdot)$, and $$ X_{t} - X_0 = \int_0^{t} \mu(X_s) \mathrm d s + \int_0^{t} >\sigma(X_s) \mathrm d B_s,$$ holds almost surely for all $t\geq 0$.
As a result of there being two different definitions of solutions, you have to distinguish two different types of "uniqueness"; If you look at pathwise uniqueness, that is that if $X_t$ and $Y_t$ are two solutions to the SDE, then $$\mathbb P( |X_1 - Y_t| = 0 ) = 1 \quad \text{for all} \quad t\geq 0,$$ then you need to consider strong solutions, as weak solutions are in general defined on different probability spaces (they require different Brownian motions to be defined) and as such there is no common probability measure to compute $\mathbb P(|X_t - Y_t| = 0)$. (Note that weak solutions are never unique, even when they are defined on the same probability space.)
In conclusion, if you want pathwise uniqueness of the solution you need to look at strong solutions and, hence, you have to have an initial condition.
See also this beautiful answer on the details about weak and strong solutions.
