NOTE.- Looking for information on Egyptian fractions, I find that, contrary to what I believed, this interesting property of unit fractions has already been established previously. So I edit this post giving the easy solution before deleting it tomorrow.
We consider the diophantine equation $$\dfrac1a = \dfrac{1}{x_1} + .... + \dfrac{1}{ x_n}\tag 1$$ A trivial solution comes from $\dfrac1a=\dfrac{n}{na}$ but finding a non-trivial solution could be very difficult even when $n$ is small. However, there is a very easy deductible way that leads us to state the following problem:
Let $a$ and $n$ be arbitrary positive integers. Prove that equation $(1)$ has at least $n-1$ non-trivial solutions $(a_1, ...., a_n)\in\mathbb N^n$.
Solution.- We have the identity $\dfrac{1}{a}=\dfrac{1}{a+1}+\dfrac{1}{a(a+1)}$ . By iteration there are 2 distinct ways to solve the equation for three unit fractions, 3 ways for four, 4 ways for five and so on.
Your solution works, but obviously, there are in general much more than $n-1$ solutions to your equation. As TheSilverDoe's pointed in the comment, you can generate a lot of solutions just using the property that $$1 = \frac{1}{2} + \frac{1}{2}$$
Here's a sketch on how to find easily a lot a solutions. First note that the $a$ is useless : indeed, if you can prove the property for $a=1$, then multiplying all the $x_i$'s by $a$, you get the property for $a$. So it is enough to prove it for $a=1$.
For $n=2$, you have the obvious decomposition $$1 = \frac{1}{2} + \frac{1}{2}$$ and for $n=3$, you have the two decompositions $$1 = \frac{1}{2} + \frac{1}{4} + \frac{1}{4} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3}$$
Now, let me explain how you can generate a lot of solution, by induction. If you have a certain solution for $n$, let's say that $$1 = \frac{1}{x_1} + ... + \frac{1}{x_n}$$
you have to ways to generate a solution for $n+1$. The first one is two divide one of the term in two equal parts (so you can generate $n$ new solutions, even if they can be the same) :
$$1 = \frac{1}{x_1} + ...+ \frac{1}{2x_i}+\frac{1}{2x_i}+ ... + \frac{1}{x_n}$$
The second one is to divide all the terms by $2$, writing $$1 = \frac{1}{2} +\frac{1}{2x_1} + ... + \frac{1}{2x_n}$$
You can check that this algorithm lets us construct a very high number of solutions, and can answer your original question. Moreover, if I come back to the original problem with the $a$, all the solutions constructed this way involve only multiples of $a$ (whereas your solution does not involve necessarily multiples of $a$) : that means that there are much more solutions than the one you constructed.
Asking exactly (or estimate precisely) how many solutions the equation has could be an much more interesting problem, in my opinion.