This is only a partial answer, but it's a bit too large to put in a comment. I will use $\mathbb{N}$ to be the set of positive integers, which includes zero.
We can represent the bowls as a couple of positive integers (can be zero) $(m, n)$. The first integer is the number of pebbles from the bowl you're picking from. The second integer is the number of pebbles in the other bowl. Now, if $m$ is even we go to $(m/2, n + m/2)$, where we are still going to pick from the first bowl. If $m$ is odd we go to $(n + (m+1)/2, (m-1)/2)$ (we swapped the two bowls, so we always pick pebbles from the first bowl. Abstractly we can model this as the following function:
$$T:\mathbb{N} \times \mathbb{N} \rightarrow \mathbb{N} \times \mathbb{N}:(m,n) \mapsto \begin{cases} \left(\frac{m}{2}, n + \frac{m}{2}\right) & \text{if }m\text{ is even} \\ \left(n+\frac{m+1}{2}, \frac{m-1}{2}\right) & \text{if }m\text{ is odd}\end{cases}$$
Your question can now be translated to the question:
For which $(m, n)$ there exists a $k \in \mathbb{N}$ such that $T^k(m,n)=(1, m+n-1)$, where $T^k$ is $T$ applied $k$ times and $T^0(m,n)=(m,n)$.
Properties of $T$:
The map $T$ is surjective.
Take $m,n \in \mathbb{N}$. If $n \geq m$, then $T(2m, n-m)=(m,n)$. If $n
The map $T$ is injective. Let $m,n,k,l \in \mathbb{N}$ such that $T(m,n)=T(k,l)$. If both $m$ and $k$ are even, then $(m,n)=(k,l)$. If both $m$ and $k$ are odd, then $(m,n)=(k,l)$. Without loss of generality we now can assume that $m$ is even and $k$ is odd. Since $T(m,n)=T(k,l)$ we get that
$$\begin{cases} \frac{m}{2}=l + \frac{k+1}{2} \\ n+\frac{m}{2}=\frac{k-1}{2}\end{cases}$$
Since $n \geq 0$ we get that $\frac{k-1}{2} =n +\frac{m}{2} \geq \frac{m}{2} = l + \frac{k+1}{2}$, hence $l \leq -1$, which is a contradiction. This proves that $T$ is injective, hence $T$ is bijective.
The orbit of $(m,n)$ is a subset of $\{(k,l) \in \mathbb{N} \times \mathbb{N} \mid k+l=m+n\}$, hence finite. This means that there is a $k \in \mathbb{N}$ such that $T^k(m, n)=(m,n)$. This $k$ is smaller or equal to the size of its orbit, hence $k \leq m+n+1$. This also means that if $T^k(m,n) = (1, m+n-1)$, then $k \leq m+n$. It is easy to see that
$T^k(m,n)=(1, m+n-1)$ for some $k \in \mathbb{N}$ if and only if $(1, m+n-1)$ is in the orbit of $(m,n)$.
We can calculate the orbit of $(1,m+n-1)$ by keep applying $T^{-1}$ ($T$ is also possible). In our proof of the surjectivity we basically found the inverse
$$T^{-1}:\mathbb{N} \times \mathbb{N} \rightarrow \mathbb{N} \times \mathbb{N}:(m,n) \mapsto \begin{cases} \left(2m, n -m\right) & \text{if }m \leq n \\ \left(2n+1, m-n-1\right) & \text{if }m > n\end{cases}$$
Let $k$ be an integer such that $2^k \leq n+m$, i.e. $2^{k-1} \leq n+m-2^{k-1}$, then $T^k(1, n+m-1)=T^{k-1}(2, n+m-2)=\cdots=(2^k, n+m-2^k)$. If $K$ is the biggest integer such that $2^K \leq n+m$, i.e. $2^K \leq n+m < 2^{K+1}$, then we get that $(2n+2m-2^{K+1} + 1, 2^{K+1}-n-m-1)$ is in the orbit of $(1, n+m-1)$.
To get to the computational side of this, you could start with $(1, m+n-1)$ and keep applying $T^{-1}$ (you can also use $T$) until you get $(1, m+n-1)$ back. All intermediate values are the pairs $(k,l)$ such that $k+l=m+n$ and $T^d(k,l)=(1, m+n-1)$ for some $d \in \mathbb{N}$.
I know it's just a partial answer, but I hope I've been able to help you.
Update
As suggested by Evangelos Bampas:
This is a plot such that a square at coordinate $(m,n)$ is coloured black if $T^k(m,n)=(1, m+n-1)$ for some $k \in \mathbb{N}$, and coloured white otherwise. You can see vertical lines corresponding to $(2^k, n)$.
Update 2
didgogns showed that $(n,n)$ (for $n > 0$) always results in win since $T^2(1, 2n-1) = (n,n)$. Now similar to this, we have that the tuples $(n, n+1)$ with $n>0$ always win. This is because $$T^2(1, 2n)=T(2n+1,0)=(n+1, n).$$
For every $n> 0$ and $\ell\geq k$ the tuple $(2^k n, (2^\ell-2^k)n)$ always wins since
$$T^{\ell-k+1}(1, 2^\ell n-1) = T^{\ell-k}(2^\ell n, 0)=(2^kn, (2^\ell-2^k)n)$$
By taking $k=0$ we get the lines (starting from the origin) above the diagonal.
