Proving that the recurrence $T(n) = 2T\left(\frac{n}{2}\right) + 1$ with $T(2) = 1$ is asymptotically $O(n)$
If you can solve a recurrence exactly, then there isn't much motivation to solve it approximately, unless the solution is much easier.
Nevertheless, here is a trick that could actually be useful. Define
$$
S(n) = T(n)+1.
$$
Then
$$
S(n) = T(n) + 1 = 2T(n/2) + 2 = 2S(n/2),
$$
and so $S(2^k) \leq 2^k S(1)$ by induction.
Let's see what this trick gives you more generally.
Application 1
Consider the recurrence $A(n) = 2A(\lfloor n/2 \rfloor) + (-1)^n$, with base case $A(1)$.
Define a new recurrence $T(n) = 2T(\lfloor n/2 \rfloor) + 1$, with base case $T(1) = A(1)$. Prove by induction that $A(n) \leq T(n)$. The technique above shows that $T(n) = O(n)$, hence $A(n) = O(n)$.
Application 2
Consider the recurrence $B(n) = 2B(\lfloor n/2 \rfloor) + \log n$, with base case $B(1)$.
Define $R(n) = B(n) + \log n$. Then
$$
R(n) = B(n) + \log n = 2B(\lfloor n/2 \rfloor) + 2\log n = 2R(\lfloor n/2 \rfloor) - 2\log \lfloor n/2 \rfloor + 2\log n.
$$
You can check that $2\log n - 2\log \lfloor n/2 \rfloor = O(1)$, and so the previous technique shows that $R(n) = O(n)$. It follows that $B(n) = O(n)$.
Application 3
Consider the recurrence $C(n) = 2C(\lceil n/2 \rceil) + 1$, with base case $C(1)$.
Defining $D(n) = C(n) + 1$ with base case $D(1) = C(1)$, we see that
$$ D(n) = 2D(\lceil n/2 \rceil). $$
Let us try to prove by induction that $D(n) \leq Kn - L$. Since $\lceil n/2 \rceil \leq n/2 + 1/2$, if the induction hypothesis holds for $\lceil n/2 \rceil$ then
$$
D(n) \leq 2K(n/2 + 1/2) - 2L = Kn + K - 2L.
$$
Choosing $L = K$, we get $D(n) \leq Kn - L$. Choosing $K = D(2)$, the induction hypothesis holds in the base case $n=2$. Altogether, we have proved $D(n) \leq K(n-1)$ (for $n \geq 2$), and so $D(n) = O(n)$, implying $C(n) = O(n)$.