When is it possible to arrive from $a$ to $b$ by this procedure? Suppose that we take two natural numbers $a$ and $b$ with $a<b$.
Then, we may try to arrive at $b$ by starting at $a$ by this set of rules:
1) To arrive at $a_1$ which is such that we have $a< a_1 \leq b$ it is only allowed that we sum $a$ with one of its divisors different from $1$. If $a_1=b$ we are done.
2) If $a_1<b$ then to arrive at $a_2$ which is such that we have $a<a_1<a_2\leq b$ we perform step 1) with $a_1$.
3) Repeat 2) until we arrive at $a_{k(a,b)}=b$
As an example, take $a=12$ and $b=100$, we have $12 \to 16 \to 20 \to 40 \to 80 \to 100$.
As an example where $\gcd (a,b)=1$ take $a=15$ and $b=22$. Then we have $15 \to 18 \to 20 \to 22$.
As an example where $\gcd (a,b)=1$ and where we cannot arrive at $b$, take $a=18$ and $b=23$.
Of course, $k(a,b)$ can be multi-valued since for the example $a=12$ and $b=100$ we also have $12\to 24\to 48\to 96\to 100$. But this question is not about behaviour of $k(a,b)$ (although it could be that its behaviour has some awesome features).
If this procedure exists for some $a$ and $b$ and $a<b$ we may say that $b$ is reachable by $a$. 

Can we find some condition(s) on $a$ and $b$ that is(are) both necessary and sufficient to guarantee that $b$ is reachable by $a$?

Since I am not sure did I explain this procedurally in a good way by 1), 2) and 3), I will describe here again the procedure in other words. It is very simple: choose natural numbers $a$ and $b$ with $a<b$. We are trying to arrive at $b$ by adding to $a$ some divisor of $a$ different than $1$. Then we arrive at $a_1=a+\alpha_1$ , where $\alpha_1 | a$. Then if we did not arrive at $b$ with that single step, we add to $a_1$ some divisor of $a_1$ different than $1$ to arrive at $a_2=a_1 + \alpha_2$, where $\alpha_2 | a_1$. We repeat this until we arrive at $b$, if possible. The question is: when is this possible?
 A: When $\gcd(a,b)>1$, then we can reach $b$ from $a$ by simply repeatedly adding $\gcd(a,b)$.
The least reachable number (LRN) from $a$ is adding its smallest prime divisor, and the greatest "arrivable" number (GAN) to $b$ is subtracting its smallest prime divisor. LRNs and GANs are always even (when defined). Since even numbers have $\gcd$ at least $2$, they are reachable when the difference between $a$ and $b$ is "large".
So the full statement is:
for $a<b$, $a$ can arrive at $b$ iff exactly one of the following holds:


*

*$a$ can arrive at $b$ in $1$ step $\iff 1<b-a=\gcd(a,b)$

*$a$ can arrive at $b$ in $2$ or more steps $\iff$ GAN of $b$ exists and is at least LRN of $a$


Sufficiency is constructive and necessity is obvious by definition of LRN and GAN.
A: Let lpf stand for least prime factor.  (Assuming gcd(a,b) is 1 and also both are larger than 1,) If b is at least as big as a plus lpf(a) plus lpf(b), then you can reach b from a.  This is because you can almost always take steps of size 2 to reach the goal.  When you can't take a step of size 2, then b must be at most lpf(a) plus lpf(b) plus a, and if b is reachable at all it must be at least that big.
Gerhard "Taking Small Steps Is Easier" Paseman, 2017.09.26.
