Your recurrence satisfies neither $T(n) = O(\log n)$ nor $T(n) = \Omega(n)$.
What does hold is that there exists an infinite sequence of values $n_k$ such that $T(n_k) \leq C\log n_k$, and another infinite sequence of values $m_\ell$ such that $T(m_\ell) \geq c m_\ell$, where $C,c>0$ are some constants.
The definitions of big O and big Omega require such inequalities to hold for all large enough $n$, rather than for infinitely many values.
A similar situation is that of partial limits and limits superior and inferior. Consider the sequence $\sin n$ (where the angle is measure in radians). For each $x \in [-1,1]$, we can find an infinite sequence $n_k$ such that $\sin n_k \to x$. These are the partial limits of the sequence. The limit superior of the sequence is $1$, and its limit inferior is $-1$.
Similarly, in your case, $\Theta(\log n)$ and $\Theta(n)$ are "partial asymptotic limits", in the sense that there is an infinite sequence $n_k$ such that $An_k \leq T(n_k) \leq Bn_k$ for some $A,B>0$, and another infinite sequence $m_\ell$ such that $C \log m_\ell \leq T(m_\ell) \leq D \log m_\ell$ for some $C,D>0$. The "asymptotic limit superior" is $O(n)$, since $T(n) = O(n)$ but $T(n)$ is not $O(f(n))$ for any $f(n) = o(n)$. Similarly, the "asymptotic limit inferior" is $\Omega(\log n)$, since $T(n) = \Omega(\log n)$ but $T(n)$ is not $\Omega(g(n))$ for any $g(n) = \omega(\log n)$.
(Showing that $T(n) = O(n)$ and that $T(n) = \Omega(\log n)$ requires an argument.)
A sequence such as $\sin n$ has no limit. Similarly, $T(n)$ has no "asymptotic limit", that is, there is no "nice" function $h(n)$ such that $T(n) = \Theta(h(n))$ (we can define "nice" in various ways, for example a logarithmico-exponential function, that is, a function which can be expressed using arithmetic operations, $\log$ an $\exp$).