Why does Coulomb's constant have units?
When the electrostatic force was originally being studied, force, mass, distance and time were all fairly well understood, but the electrostatic force and electric charge were new and exotic. In the cgs system, the charge was defined in relation to the resulting electrostatic force (it's called a Franklin (Fr) an "electrostatic unit" (esu or) sometimes a statCoulomb (statC)).
In that system, we express the force on one charged particle by another as $F_E=\frac{q_1 q_2}{r^2}$ where the unit of charge is the esu, the unit of force is the dyne and the unit of distance is the centimeter. In the MKS system (now called SI), we would write $F_E = k_e\frac{q_1 q_2}{r^2}$ where the unit of charge is the coulomb, the unit of force is the newton, and the unit of distance the meter. It would seem that if things are equivalent, then $k_e$ is indeed just a conversion factor, but things are definitely not equivalent.
A little history is probably useful at this point. In 1873, when the cgs system was first standardized, it finally made a clear distinction between mass and force. Before that, it was common to express both in terms of the same unit, such as the pound. So if you think of it, people still say things like "I weigh 72 kg" rather than "I weigh 705 N here on the surface of Earth" and they also say $1 \mathrm { kg} = 2.2\mathrm{ lb}$ confusing mass and weight (the cgs Imperial unit of mass is actually the slug).
This is important, because there is a direct analogy to the issue of units of charge and to your question about the units of $k_e$. The Franklin is defined as "that charge which exerts on an equal charge at a distance of one centimeter in vacuo a force of one dyne." The value of $k_e$ is assumed to be 1 and is dimensionless in the cgs system.
In cgs, the unit of charge, therefore, already implictly has this value of $k_e$ built in. However in the SI units, they started with Amperes and derived Coulombs from that and time ($C=It$). The resulting units of $k_e$ are a result of that choice.
So although the physical phenomenon is the same, it is the choice of units that either gives $k_e$ dimension or not.
See this paper for perhaps a little more detail on how this works in practice.
Systems of units are in some sense flexible and optional.
The relationship
$$ \text{Electrostatic force twixt two point-like charges} \propto \frac{(\text{one charge})\times (\text{the other charge})}{(\text{distance between them})^2} \tag{1}$$
is an experimental fact.
In SI, we have units for Force, distance and charge such that (1) is not dimensionally consistent with a dimensionless constant of proportionality. So, $k$ must have dimensions of $\frac{\mathrm{N} \cdot \mathrm{m}^2}{\mathrm{C}^2}$ as well as having a numeric value.
But we could do it another way. Consider the "Statcoulomb". In Gaussian units the unit of charge is defined such that Coulomb's Law has a dimensionless, unit constant of proportionality. $$ F = \frac{q_1 \, q_2}{r^2} \,.$$ This is a perfectly valid way to do physics. In essence we have folded $\sqrt{k}$ into the numerical value of each of the charges:
$$ (\text{charge})(in\; \text{Statcoulombs}) \sim \sqrt{k} \, (\text{same charge})(in\; \text{Coulombs}) \,,$$
or
$$ \sqrt{k} \,\text{Statcoulombs} \sim 1 \,\text{Coulombs} \,.$$
That makes a Statcoulomb a pretty funny unit when expressed in SI terms, but then the Coulomb is a pretty odd unit expressed in Guassian terms. Each system should be understood in it's own context.
A great many words have been spilled arguing that one set of units is better than another or vice-versa.
In my business (particle physics) it is common to work in units where $c = \hbar = 1 \,(\text{dimensionless})\;.$ This gives energy, mass, and momentum the same units (inverse distance, actually) and loses many of the checks that help young physicists keep track of the difference between these quantities, but keeps the scribbling down and simplifies the form of many equations. (By the way, cosmologist often add $G = 1\,(\text{dimensionless})$ to the mix.
The moral of the story is, 'Don't read too much significance into the units of "constants", because they depend on the system of units you chose.'
Force is a vector quantity defined mathematically as the rate of change of momentum, that is
$\vec F = \dfrac{d\vec p}{dt} \tag{1}$ where $\vec p= m\vec v$ in classical mechanics.
The unit of force in Si is "newton". One "newton" is the amount of force requires to accelerate one kilogram mass at a rate of one meter per second squared. You could make your own set of units by saying : I define one newton as the force required to accelerate 2 Kg of mass through 1m/s^2 then you will have to modify eqn 1 as $\vec F=\dfrac{1}{2}m \vec a$
In general Newton's second law can be stated as $\vec F=k \ m\vec a$ where $k$ depends upon the units of measurement.you could also say $\vec F \propto d\vec p/dt$ and $k$ appears as a constant of proportionality. It should be noted here that '$k$' is dimensionless constant.
Let our system of units be SI for the sake of simplicity and eqn 1 be valid. Let us measure the force by a spring scale
The spring obeys Hooke's law which states that the magnitude of applied force is directly proportional to the displacement of the spring, that is $F=k X$ where $x$ is the displacement and $k$ is the proportionality constant. This time $k$ is not dimensionless why it is so? it is so because force is not measured in metres it is measured in newtons. Suppose the spring is manufactured in such a way that a displacement of one metre represent one newton of force then $k$ will have $1$ magnitude. Would it be appropriate to define to define one newton = one metre so as to make $k$ dimensionless? NO! doing this will make all other equations involving '$F$' dimensionally incorrect e.g $F=kma$ here k will have to be dimensional constant.
Similarily if you define "one newton = $1 \text{Newton} = \frac{1}{1/1 \text{meter}^2 \cdot 1 \text{Coloumb}^2}$ then $F=ma$ will have to be changed to $ F=\dfrac{s^2}{\text{Kg m}} \times {1/\text{meter}^2 \cdot \text{coloumb}^2} \times \text{mass} \times \text{acceleration} $
In Coloumb's law the constant $k_e$ has units and magnitude.
$1.$ It has units to make the equation dimensionally correct.
$2.$ it is not $1$ in magnitude because the magnitude of $\dfrac{q_1q_2}{r^2}$ when $q_1$,$q_2$ and $r$ all are one the force in newtons is found to be $9\times 10^9$ in magnitude so $k$ is give this value for proper calibration. $k$ can be seen as a proportionality constant or a scaling factor or a dimensional constant or all. $k$'s magnitude is 9*10^9 only because we defined one newton as 1Kg m/s^2. If you say 9*10^9 newtons= one dgp then in dgp system $k$ will attain one magnitude.