I believe you are correct. Using conservation of linear momentum:
$$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$$
Where the mass of the person is $m_1$, his speed before he hits the wall is $u_1$, the mass of the cylindrical space ship is $m_2$ and the initial velocity of the cylinder is $u_2$. $v_1$ and $v_2$ are their respective velocities post collision.
The cylinder is at rest so $u_2 = 0$. Assuming that the person running comes to rest once he hits the wall:
$$m_1u_1 = m_2v_2$$
$$v_2 = \frac{m_1u_1}{m_2}$$
Therefore, $v_2 > 0$, so the cylinder now has some kinetic energy. It will be a very small amount of kinetic energy, because $m_2 >> m_1u_1$, but it will have kinetic energy nonetheless.
If the person were to rebound, $v_1$ would be in the negative direction, so the equation would look like this:
$$m_1u_1 = m_2v_2 - m_1v_1$$
$$v_2 = \frac{m_1(u_1 + v_1)}{m_2}$$
In this case, $v_2$ would be even greater.
So yes, in order to conserve momentum, the ship would have to have some kinetic energy, no matter how small, post collision.