If measurement causes entanglement with the observer, is any further measurement possible? When an observer measures a system, the systems wavefunction collapses and they become entangled. Does this mean that any further measurement by the observer on the system is impossible, as the system no longer has a wavefunction that is independent of the observer to collapse?
 A: "When an observer measures a system, the systems wavefunction collapses and they become entangled."
That's not quite correct. When systems interact, they become entangled. They remain entangled even if separated. When an observer observes one of the entangled systems, the joint wavefunction of both collapses, 'instantaneously', faster-than-light, in some vaguely-defined and unobservable sense. They are no longer in an entangled state - they are each in a pure state - the pure states they each collapse to are correlated, but now in an ordinary, classical, non-spooky manner. For interpretations of quantum mechanics that posit wavefunction collapse, that is.
[For other interpretations, like the Everett interpretation, systems that interact become entangled. They remain entangled even when separated. When an observer observes one part, the observer becomes entangled with both but no wavefunction collapse happens. Nothing travels faster than light. Nothing happens to the other unobserved part of the system. When it too is observed and the observers compare notes, the observers end up entangled in such a way that each only perceives a partner making the consistent, correlated observation.]
Either way, they're entangled first, before the measurement, which is why the outcomes are correlated.
"Does this mean that any further measurement by the observer on the system is impossible, as the system no longer has a wavefunction that is independent of the observer to collapse?"
In either type of interpretation, further measurements can certainly be made. If the observer makes exactly the same sort of measurement (same observable), and does so fast enough that the state hasn't evolved, she gets the same result. (Collapse the second time around has no effect - the first collapse left it in a pure state for which the measurement gives the same outcome with 100% probability.) If the observer makes a different measurement but for a commuting observable, this gives a different result but the pure eigenstate is undisturbed. Going back to the first observable gives the same result you got originally. If the observer instead makes a complementary, non-commuting sort of measurement (e.g. momentum instead of position) this tends to scramble the entanglement, and then another measurement of the first sort gives a new random outcome.
A: 
When an observer measures a system, the systems wavefunction collapses... Does this mean that any further measurement by the observer on the system is impossible...

No. You can keep measuring at much as you want. (Please see my update below. I am discussing measurement of a pure state. This is textbook quantum mechanics.)
The "collapse" means that the wavefunction is projected onto the eigenstates of the measured operator that correspond to the measured eigenvalue. The "collapsed" state is still a wavefunction, just a different wavefunction.
For example, if you measure that the position is a $\vec x$, the wavefunction "collapses" to state that is typically denoted by $|\vec x\rangle$. This ket is still a wavefunction and its further evolution is still governed by whatever Hamiltonian/Schrodinger equation previously governed the evolution.

Update
To clarify, I am referring to measurement of a pure state, which is a well-known concept in Quantum Mechanics. See Volume 1, Chapter V, Section 13 of Messiah's Quantum Mechanics textbook, an excerpt of which is summarized below.
Supposed the wavefunction is given by:
$$
\Psi = \sum_p\Psi_p\;,
$$
where
$$
\Psi_p = \sum_r c_{pr}\phi_{pr}\;
$$
Here, $p$ means the part of the wave function that corresponds to the eigenvalue $a_p$ of the observable $A$.
The probability to measure $a_i$ is:
$$
\sum_r |c_{ir}|^2\;.
$$
After an ideal measurement, if the value $a_i$ is measured the wavefunction "collapses" to:
$$
\Psi_i = \sum_r c_{ir}\phi_{ir}\;.
$$
This is really not debatable. So, I'm not sure why this is downvoted...
If the measurements is not "ideal" the wavefunction is still projected onto the same subspace, it is just that the coefficients $c_{ir}$ can be modified. The wavefunction still lives in the projected subspace after measurement:
$$
\Psi \to \Psi' = \sum_{r}b_r \phi_{ir}
$$
