Equality by iteratively applying $(a,b)\rightarrow [(a+1,2b)\text{ or }(2a,b+1)]$? 
I play a game starting with $2$ positive integers $a$ and $b$. At each step of the game I can double one of the integers and add $1$ to the other integer. Is there always a procedure for any starting pair $(a,b)$ so that eventually my two numbers are equal?

An example of a successful procedure starting from the pair $(1,4)$:
$$(1,4)\rightarrow(2,5)\rightarrow(3,10)\rightarrow(6,11)\rightarrow(12,12)$$
The original question did not restrict to positive integers, but I have solved the cases where at least one of the starting integers is nonpositive:
Case (1): Both are nonpositive. Solution: Continually add $1$ to one of the numbers until it is $0$. Then keep doubling the $0$ while adding $1$ to the other number until it is also $0$.
Case(2): One of the numbers is nonpositive. Solution: Add $1$ to the nonpositive number until it is $0$. Next double $0$ and add $1$ to the other number until the other number is in the form $\dfrac{2^n-n}{2}$ for some $n$. Now we can perform the procedure
$$\left(0,\frac{2^n-n}{2}\right)\rightarrow(1,2^n-n)\rightarrow(2,2^n-n+1)\rightarrow(4,2^n-n+2)\rightarrow\dots\rightarrow(2^n,2^n)$$
I remain stuck on the case in which both starting integers are positive (though I've been able to find a valid procedure for every pair I've actually tried).
 A: Call a pair equalizable if there is a procedure ending in equal numbers.
Clearly any pair of the form $(k,4k)$ is equalizable:
$$
(k,4k)\to (2k,4k+1)\to(2k+1,8k+2)\to(4k+2,8k+3)\to(8k+4,8k+4).
$$
From this we deduce that any pair $(k,2k)$ is equalizable:
$$
(k,2k)\to (k+1,4k)\to(2k+2,4k+1)\to(4k+4,4k+2)\to(8k+8,4k+3)\to (16k+16,4k+4).
$$
Hence any pair that has a procedure leading to $(m,m-1)$ or $(m,m-2)$, is equalizable; take $(m,m-1)\to (2m,m)$ and $(m,m-2)\to (2m,m-1)\to (4m,m)$.
From this we deduce that any pair $(k,8k)$ is also equalizable:
$$
(k,8k)\to (4k,8k+2)\to(4k+1,16k+4)\to(16k+4,16k+6).
$$
Since the difference is 2, it is equalizable. 
Hence, by the same argument as above, any pair that has a procedure leading to $(m,m-3)$, is equalizable.
Now we prove that any pair $(k,16k)$ is also equalizable:
$$
(k,16k)\to (4k,16k+2)\to(4k+1,32k+4)\to(32k+8,32k+7).
$$
Since the difference is 1, it is equalizable. Hence, by the same argument as above, any pair that has a procedure leading to $(m,m-4)$, is equalizable.
Now we prove that any pair $(k,32k)$ is also equalizable:
$$
(k,32k)\to (8k,32k+3)\to(8k+1,64k+6)\to(64k+8,64k+9).
$$
Since the difference is 1, it is equalizable. Hence, by the same argument as above, any pair that has a procedure leading to $(m,m-5)$, is equalizable.
Now we proceed by induction:
Assume that we have proved that any pair of the form $(k,k \cdot 2^j)$ is equalizable, for $j=1,\dots,n-1$. Then it follows as above, that any pair $(s,t)$ with $|s-t|<n$ is equalizable. Below we will prove that for any $n>5$, there exists $r$ with $1<r<n$ such that
$$
|2^{n+1-r}-(n+r+1)|<n.
$$
Now we prove that $(k,k\cdot 2^n)$ is equalizable:
$$
(k,k\cdot 2^n)\to (k\cdot 2^r,k\cdot 2^n+r)\to (k\cdot 2^r+1,k\cdot 
2^{n+1}+2r)\to 
$$
$$
\small \to(k\cdot 2^{n+1}+2^{n+1-r},k\cdot 2^{n+1}+2r+(n+1-r))=(k\cdot 2^{n+1}+2^{n+1-r},k\cdot 2^{n+1}+n+1+r)
$$
For the difference we have $|2^{n+1-r}-(n+r+1)|<n$, hence the pair $(k,k\cdot 2^n)$ is equalizable.
Finally we prove that for any $n>5$, there exists $r$ with $1<r<n$ such that
$$
|2^{n+1-r}-(n+r+1)|<n.
$$
We define the decreasing function $f(k):=2^{n+1-k}-(n+k+1)$ which fulfills $f(1)>0$ and $f(n-2)=8-(2n-1)<0$. Hence there exists $k\le n-3$ with $f(k)\ge 0$ and $f(k+1)<0$. Then
$$
|f(k)|+|f(k+1)|=2^{n-k}+1=f(k+1)+n+k+3<2n
$$
This proves the claim, since then either $|f(k)|<n$ or $|f(k+1)|<n$.
