First a comment on notation. It seems that you are confusing what it means to add vectors. Remember that the $x$ you have inside a bra or a ket is just a label, not a real number to be added to other real numbers. What it means to have a position eigenvector $|x\rangle$ is that
So by the definition of addition you have presented we should have
But this is very confusing. Why would being in a superposition between $x_1$ and $x_2$ have anything to do with the completely different and uncorrelated point $x_1+x_2$?
When you start doing this when mixing diferent basis (i.e. position and momentum) you're bound to have a bad time.
Now for more pressing matters. There's nothing stopping you from being in a superposition between position states and momentum states, or to define vectors that are a superposition of those. You just have to be careful with normalizations. Let's say we want a vector that represents a superposition of a state $|\Psi\rangle$, with position wavefunction $\psi_0(x)$, and a state $|\Phi\rangle$, with momentum eigenfunction $\phi_0(p)$. We can then write
Here we have that these are honest-to-goodness physical states if
$$\int dx\,|\psi_0(x)|^2=\int dp\,|\phi_0(p)|^2=1.$$
Now let's look at what happens when we act on your state $|\mathcal S(t)\rangle$ with the bra $\langle\Psi|+\langle\Phi|$, the superposition of the vectors we've just defined. Linearity and the distributive property hold just fine,
$$(\langle\Psi|+\langle\Phi|)|\mathcal S(t)\rangle=\langle\Psi|\mathcal S(t)\rangle+\langle\Phi|\mathcal S(t)\rangle$$
$$(\langle\Psi|+\langle\Phi|)|\mathcal S(t)\rangle=\int dx\,\psi^*_0(x)\langle x|\mathcal S(t)\rangle+\int dp\,\phi^*_0(p)\langle p|\mathcal S(t)\rangle$$
$$(\langle\Psi|+\langle\Phi|)|\mathcal S(t)\rangle=\int dx\,\psi^*_0(x)\Psi(t,x)+\int dp\,\phi^*_0(p)\Phi(t,p),$$
which makes perfect sense and you can check that the dimensions are correct. The main issue with the dimensions is that, because $x$ and $p$ are continuous variables, they have no true eigenstates. State vectors should be dimensionless, and that does not hold for things of the type $|x\rangle$ and $|p\rangle$. But if you combine them with valid wavefunctions like $\psi_0$ and $\phi_0$, they combine to dimensionless, physically meaningful vectors.