Does spin entanglement imply position entanglement?
No. (In the sense that I can write down a valid multi-electron wavefunction that is entangled with respect to spin, but is not entangled with respect to space.)
My question is whether two electrons can be entangled only with respect to their spins but not with respect to some other observable, such as position.
Yes, it is possible. If the spin-entangled state is anti-symmetric then the spatial part can be an unentangled direct product (symmetric) state.
For simplicity let's consider just two electrons. Electrons are fermions, so at the end of the day the overall wavefunction of both electrons must be anti-symmetric:
$$
\Psi(\vec r_1,s_1;\vec r_2,s_2) = -\Psi(\vec r_2,s_2;\vec r_1,s_1)\;,
$$
where $\vec r$ is the spatial index and $s$ is the spin index.
I initially believed that spin-entanglement doesn't entail position or momentum entanglement. But if particles $a$ and $b$ have a composite state given by $|\Psi\rangle_{ab} = \frac{1}{\sqrt2}(|1\rangle|0\rangle + |0\rangle|1\rangle)$, where $|0\rangle$ and $|1\rangle$ are distinct z-spin eigenstates, then...
In your example there must be some elided/implicit spatial part of the wave function, since your wave function $|\Psi\rangle_{ab}$ alone is symmetric. (It is not uncomment to elide such information. For example, in some cases we can even treat electrons as distinguishable (approximately, of course).)
To be more complete, we can re-write your spin-only wave-function as a full wavefunction (spin and space) like:
$$
\Psi(\vec r_1,s_1;\vec r_2,s_2)=\Phi(\vec r_1, \vec r_2)\frac{1}{\sqrt{2}}\left(\delta_{s_1,0}\delta_{s_2,1}+\delta_{s_2,0}\delta_{s_1,1}\right)\;,
$$
where we must have
$$
\Phi(\vec r_1, \vec r_2) = -\Phi(\vec r_2, \vec r_1)\;,
$$
i.e., $\Phi$ is anti-symmetric, and
where $\Phi$ is normalized as
$$
\int d^3r d^3r'|\Phi(\vec r, \vec r')|^2=1\;.
$$
In your example, the spatial part of the wave function certainly can not be a simple direct product, since that would either be symmetric or have no simple exchange symmetry property.
But, if spin state is anti-symmetric, then there is no necessary requirement that the spatial part is entangled, since a simple symmetric direct product of spatial wavefunctions could be used.
Thus, the general answer is: No. Spin entanglement does not always imply spatial entanglement.
For example, the example state
$$
\Psi_{ex}(\vec r_1, s_1, \vec r_2, s_2)=\phi_{n\ell m}(\vec r_1)\phi_{n\ell m}(\vec r_2)\frac{1}{\sqrt{2}}\left(\delta_{s_1,0}\delta_{s_2,1}-\delta_{s_2,0}\delta_{s_1,1}\right)\;,
$$
is entangled in spin (it is the spin singlet) and is unentangled in space (it is a direct product of two orbitals with the same orbital eigenvalues). And the full state is anti-symmetric under interchange of the two electrons, as required by the rules of many-body quantum mechanics.
If so, however, what about their mass or charge?
The mass and charge are both proportional to the number operator, which is always two in these example cases.
$$
\hat N = 2\;.
$$
(I.e., $\hat N$ is proportional to the identity operator for two-particle states, and the proportionality constant is 2.)
So, we also have
$$
\hat Q = 2e\;,
$$
where $e$ is the electronic charge and
$$
\hat M = 2m_e\;,
$$
where $m_e$ is the electronic mass.
