The theorem of existence and uniqueness is: Let y ′ + p ( x )...

The existence and uniqueness theorem for first-order linear differential equations can be stated as follows. Suppose that P and Q are continuous on the open interval I. If a and b are any real numbers, then there is a unique function y=f(x) satisfying the initial-value problem $y^{\prime }+P(x)y=Q(x)$ with f(a)=b on the interval I. With regard to your question, the important point is that a and b are arbitrary real numbers and the unique solution f to the differential equation satisfies f(a)=b for every choice of a and b. Since every first-order linear differential equation satisfying the constraints of the theorem has a solution satisfying f(a)=b, there is no case in which such an equation has no solution satisfying f(a)=b.
If we look at the simpler case of homogeneous first-order linear differential equations of the form $y^{\prime }+P(x)y=0$, where P is continuous on the open interval I, we can directly verify that for every choice of a and b, the function $f(x)=b{e}^{-A(x)}$ where $A(x)={\int }_a^{x}P(t)dt$ is a solution to $y^{\prime }+P(x)y=0$. Now letting g be an arbitrary solution of $y^{\prime }+P(x)y=0$, we establish uniqueness by showing that $g(x)e^{A(x)}=b$. Differentiating, we see that $h(x)=g(x)e^{A(x)}$ is constant on the interval I. But h(a)=b, so we must have h(x)=b. This demonstrates that g=f. Notice that the choice of a and b does not affect the existence or uniqueness of solutions. Can you verify this in the case of non-homogeneous first-order linear differential equations of the form $y^{\prime }+P(x)y=Q(x)$