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.
Physics is independent of our choice of units
And for something like a length plus a time, there is no way to uniquely specify a result that does not depend on the units you choose for the length or for the time.
Any measurable quantity belongs to some set $\mathcal{M}$. Often, this measurable quantity comes with some notion of "addition" or "concatenation". For example, the length of a rod $L \in \mathcal{L}$ is a measurable quantity. You can define an addition operation $+$ on $\mathcal{L}$ by saying that $L_1 + L_2$ is the length of the rod formed by sticking rods 1 and 2 end-to-end.
The fact that we attach a real number to it means that we have an isomorphism
$$
u_{\mathcal{M}} \colon \mathcal{M} \to \mathbb{R},
$$
in which
$$
u_{\mathcal{M}}(L_1 + L_2) = u_{\mathcal{M}}(L_1) + u_{\mathcal{M}}(L_2).
$$
A choice of units is essentially a choice of this isomorphism. Recall that an isomorphism is invertible, so for any real number $x$ you have a possible measurement $u_{\mathcal{M}}^{-1}(x)$. I'm being fuzzy about whether $\mathbb{R}$ is the set of real numbers or just the positive numbers; i.e. whether these are groups, monoids, or something else. I don't think it matters a lot for this post and, more importantly, I haven't figured it all out.
Now, since physics should be independent of our choice of units, it should be independent of the particular isomorphisms $u_Q$, $u_R$, $u_S$, etc. that we use for our measurables $Q$, $R$, $S$, etc. A change of units is an automorphism of the real numbers; given two units $u_Q$ and $u'_Q$, the change of units is
$$
\omega_{u,u'} \equiv u'_Q \circ u_Q^{-1}$$
or, equivalently,
$$
\omega_{u,u'} \colon \mathbb{R} \to \mathbb{R} \ni \omega(x) = u'_Q(u_Q^{-1}(x)).
$$
Therefore,
\begin{align}
\omega(x+y) &= u'_Q(u_Q^{-1}(x+y)) \\
&= u'_Q(u_Q^{-1}(x)+u_Q^{-1}(y)) \\
&= u'_Q(u_Q^{-1}(x)) + u'_Q(u_Q^{-1}(y)) \\
&= \omega(x) + \omega(y).
\end{align}
So, since $\omega$ is an automorphism of the reals, it must be a rescaling $\omega(x) = \lambda x$ with some relative scale $\lambda$ (As pointed out by @SeleneRoutley, this requires the weak assumption that $\omega$ is a continuous function -- there are everywhere discontinuous solutions as well. Obviously units aren't useful if they are everywhere discontinuous; in particular, so that instrumental measurement error maps an allowed space of $u_\mathcal{M}$ into an interval of $\mathbb{R}$. If we allow the existence of an order operation on $\mathcal{M}$, or perhaps a unit-independent topology, this could be made more precise.).
Consider a typical physical formula, e.g.,
$$
F \colon Q \times R \to S \ni F(q,r) = s,
$$
where $Q$, $R$, and $S$ are all additive measurable in the sense defined above. Give all three of these measurables units. Then there is a function
$$
f \colon \mathbb{R} \times \mathbb{R} \to \mathbb{R}
$$
defined by
$$
f(x,y) = u_S(F(u_Q^{-1}(x),u_R^{-1}(y)).
$$
The requirement that physics must be independent of units means that if the units for $Q$ and $R$ are scaled by some amounts $\lambda_Q$ and $\lambda_R$, then there must be a rescaling of $S$, $\lambda_S$, such that
$$
f(\lambda_Q x, \lambda_R y) = \lambda_S f(x,y).
$$
For example, imagine the momentum function taking a mass $m \in M$ and a velocity $v \in V$ to give a momentum $p \in P$. Choosing $\text{kg}$ for mass, $\text{m/s}$ for velocity, and $\text{kg}\,\text{m/s}$ for momentum, this equation is
$$
p(m,v) = m*v.
$$
Now, if the mass unit is changed to $\text{g}$, it is scaled by $1000$, and if the velocity is changed to $\text{cm/s}$, it is scaled by $100$. Unit dependence requires that there be a rescaling of momentum such that
$$
p(1000m,100v) = \lambda p(m,v).
$$
This is simple -- $10^5 mv = \lambda mv$ and so $\lambda = 10^5$. In other words,
$$
p[\text{g} \, \text{cm/s}] = 10^5 p[\text{kg} \, \text{m/s}].
$$
Now, let's consider a hypothetical situation where we have a quantity called "length plus time", defined that when length is measured in meters and time in seconds, and "length plus time" in some hypothetical unit called "meter+second", the equation for "length plus time" is
$$
f(l,t) = l + t.
$$
This is what you've said - $10 \text{ m} + 5 \text{ s} = 15 \text{
“m+s”}$. Now, is this equation invariant under a change of units? Change the length scale by $\lambda_L$ and the time scale by $\lambda_T$. Is there a number $\Lambda$ such that
$$
f(\lambda_L l, \lambda_T t) = \lambda_L l + \lambda_T t
$$
is equal to
$$
\Lambda f(l,t) = \Lambda(l+t)
$$
for all lengths and times $l$ and $t$? No! Therefore, this equation $f = l + t$ cannot be a valid representation in real numbers of a physical formula.
Best Answer
Dimensional analysis is used when you're trying to figure out how a certain set of parameters (your "inputs") can be combined to yield a quantity with a particular set of units (your "output"). It only works under the following assumptions:
You know what all of your inputs are, and the list of inputs is finite;
There are no redundancies in your list of inputs (i.e. each input has different units), or, if there are redundancies, there must be additional information that constrains the behavior of the redundant inputs (e.g. "the Coulomb force must involve both $Q_1$ and $Q_2$");
Each of these quantities are expressed in units that are compatible with each other and with your output (in practice, this means that each of them should be expressible in only SI base units);
Any constant factors are assumed to be pure numbers, and
The number of additive terms is constrained to be finite.
If these assumptions are satisfied, dimensional analysis will yield a set of possible combinations of inputs ("formulas") that will have the same units as your output. If you're lucky, there will only be one; if you're not, there will be a few, which serve as arguments for an arbitrary function (this follows from the Buckingham Pi theorem).
Let's try this procedure on your two examples. In the first one, we have a pendulum. Our output is the period, which has units of time ($T$). The inputs are:
The length of the pendulum rod, which has units of length ($L$);
The local gravitational acceleration, which has units $\frac{L}{T^2}$, and
The mass of the pendulum, which has units of mass ($M$) (since we don't know a priori that it can't be an input, we include it as a possible input).
Using dimensional analysis, we can constrain the possible forms for the formula using the following equation and solving for the powers $a$, $b$, and $c$:
$$T=L^a\left(\frac{L}{T^2}\right)^bM^c$$
where $k$ is some unitless proportionality constant. Equating the powers of $L$, $T$, and $M$ on the left-hand side with their powers on the right-hand side:
$$0=a+b\quad\quad\quad 1=-2b\quad\quad\quad 0=c$$
Solving this system gives you $a=\frac{1}{2}$, $b=-\frac{1}{2}$, $c=0$. So, even though we didn't know that the period didn't depend on mass before, we have just proven that this is the case, assuming that our list of inputs was complete. Substituting the powers back into the original expression, we get an expression for the period $\tau$:
$$\tau=k \ell^{1/2}g^{-1/2}m^0=k\sqrt{\frac{\ell}{g}}$$
Since the solution to the linear system above is unique, there is only one term.
For Coulomb's law, we must use a system in which the units of each quantity are compatible; as such, we use Gaussian electromagnetic units, in which charge has units of $M^{1/2}L^{3/2}T^{-1}$ (i.e. units of g$^{1/2}$ cm$^{3/2}$/s). The units of the separation are, of course, $L$. In this case, we have three inputs: the two charges $Q_1$ and $Q_2$, which both have the above units, and the separation between the charges. Our output is force, which has units of $MLT^{-2}$ (i.e. g cm/s$^2$). Setting up dimensional analysis again:
$$MLT^{-2}=(M^{1/2}L^{3/2}T^{-1})^a(M^{1/2}L^{3/2}T^{-1})^bL^c$$
Solving for the powers:
$$1=\frac{a}{2}+\frac{b}{2}\quad\quad\quad 1=\frac{3a}{2}+\frac{3b}{2}+c\quad\quad\quad -2=-a-b$$
This system is degenerate, and gives you one free parameter, so $(a,b,c)=(a,2-a,-2)$. As such, the abstract general form of the force that you get with dimensional analysis is
$$F=f\left(\left\{k_a\frac{Q_1^aQ_2^{2-a}}{r^2}\right\}_{a\in\mathbb{R}}\right)$$
for some arbitrary function $f$. Note that we immediately get the $1/r^2$ nature of the force just through dimensional analysis. It is also very easy to experimentally eliminate all but one of these possible arguments, by invoking one piece of additional information, that can easily be gleaned from experiment: exchange symmetry. The force on two charges does not change if you swap the two charges with each other. This means that the powers on $Q_1$ and $Q_2$ must be equal. As such, the only possible powers are $(a,b,c)=(1,1,-2)$. This additional empirical information eliminates the redundancy in this dimensional analysis, and so we arrive at the correct formula:
$$F=k\frac{Q_1Q_2}{r^2}$$
This all comes with one major caveat: if there's an input that you don't know about and don't include, then you could get an entirely different formula. For example, let's look at the pendulum again. Let's assume that in this case, the rod is very slightly elastic, with an effective spring constant $K$ that has the usual units of N/m, or abstractly $MT^{-2}$. Now, redoing the dimensional analysis yields:
$$T=L^a\left(\frac{L}{T^2}\right)^b M^c \left(\frac{M}{T^2}\right)^d$$
Which yields the following system:
$$0=a+b\quad\quad\quad 1=-2b-2d\quad\quad\quad 0=c+d$$
Note that there are 4 variables and 3 equations, so there is 1 free parameter, which I will take to be $c$. As such, we solve the system to obtain $(a,b,c,d)=(-c+1/2,c-1/2,c,-c)$, which gives us an abstract general formula:
$$\tau=f\left(\left\{ k_c\left(\frac{\ell}{g}\right)^{1/2-c}\left(\frac{m}{K}\right)^c\right\}_{c\in\mathbb{R}}\right)$$
again for some arbitrary function $f$. Now, if $c$ is nonzero, then the period of a slightly elastic pendulum does depend on the mass of the pendulum, and the particular way that it does must be measured.
But now let's consider the limit of very slight elasticity (i.e. the limit of large $K$). Equivalently, suppose $\tau$ varies slowly enough with $m/K$ that $\log\tau$ vs. $\log(m/K)$ is well-approximated by a line. This means that there is only one nonzero term in the above set of arguments, since a linear log-log plot corresponds to power-law behavior. This simplified our expression considerably:
$$\log\tau=\log k+\left(\frac{1}{2}-c\right)\log\left(\frac{\ell}{g}\right)+c\log\left(\frac{m}{K}\right)$$
Therefore, we have reduced a complicated physical task (finding the period of a pendulum with a slightly elastic rod) into the much easier task of empirically finding the two constants $k $ and $c $.