Master Theorem on oscillating function
Yes, you can.  You are defining the recurrence $T'$, so define $T'(1) = T(1)$.  Then, since $g(n)$ satisfies the regularity conditions, the Master theorem tells you that $T'(n) = \Theta(g(n))$.  Note that the Master theorem doesn't depend on the choice of $T'(1)$; it is a constant, and you get the same asymptotic bound for $T'(n)$ regardless of which constant you choose.

Now, you can prove by induction that $T(n) \le T'(n)$ holds for all $n$.  This is a simple induction.  The base case holds because you defined $T'$ that way.  The inductive step holds because $f(n) \le g(n)$ for all $n$.

Once you've proven that $T(n) \le T'(n)$ for all $n$, you are entitled to conclude that $T(n) = O(T'(n))$ and thus $T(n) = O(g(n))$.