Induction could lead you to the answer.
The equation is :

$$
frac 1 x_1 + frac 1 x_2 + dots + frac 1 x_n + frac 1 x_1 x_2 dots x_n = 1
$$
Case $ n = 0 $: the empty set solves the equation as an empty product is 1

Case $ n = 1 $: the obvious solution is $ x_1 = 2 $.

Case $ n = 2 $: a bit more difficult, but you can find $ x_1 = 2, x_2 = 3 $.
Doing this, I noticed one thing: assuming that you solved the $ (n-1) $-th equation, you can pick $ x_n $ so that $ + frac 1 x_n $ in the first part of the equation compensates the factor $ frac 1 x_n $. Let’s check.

Case any $ n $: assuming that $ x_1, dots x_n $ solves the equation, we require $ x_n+1 $ so that

$$
frac 1 x_n + frac 1 x_2 + dots + frac 1 x_n+1 + frac 1 x_1 x_2 dots x_n+1 = frac 1 x_1 + frac 1 x_2 + dots + frac 1 x_n + frac 1 x_1 x_2 dots x_n
$$

Removing identical summands:

$$
frac 1 x_n+1 + frac 1 x_1 x_2 dots x_n+1 = frac 1 x_1 x_2 dots x_n
$$
Multiplying tops by $ x_1 x_2 dots x_n+1 $ :
$$
x_1 x_2 dots x_n + 1 = x_n+1
$$

Solved!

A start, on my phone.

Assume $j<k<l<m<n.$

Then j=2 or 3 because 1 makes the sum too large and 4 makes it too small.

Therefore the left without 1/j is 1/2 or 2/3.
You can get a tree of possiblities by continuing in this way.

Another tack:

Clear fractions to get

$klmn+jlmn+jkmn+jkln+jklm+1=jklmn$

or

$klmn+j(...)+1=jklmn$

or

$j(klmn-...)=klmn+1$.

Therefore $j|(klmn+1)$
(and similarly for the others)
so that j is relatively prime to the others.

Therefore all the variables are
pairwise relatively prime.

I'll leave it at this since
that's all I can think of
lying in bed.

Unless I've made a mistake, the solutions (up to permutation) are

[2, 3, 7, 43, 1807]

[2, 3, 7, 47, 395]

[2, 3, 11, 23, 31]

Maple code:

f:= proc(S) local R;
R:= map(t -> 1/t, S);
convert(R,`+`) + convert(R,`*`)
end proc:
for jj from 2 to 3 do
 for kk from jj+1 while f([jj,kk,kk+1,kk+2,kk+3]) >= 1 do
 lmin:= floor(solve(1/jj+1/kk+1/l=1));
 for ll from max(kk+1,lmin) while f([jj,kk,ll,ll+1,ll+2]) >= 1 do
 if 1/jj+1/kk+1/ll >= 1 then next fi;
 for mm from max(ll+1,floor(solve(1/jj+1/kk+1/ll+1/m=1))) while f([jj,kk,ll,mm,mm+1]) >= 1 do
 nn:= solve(f([jj,kk,ll,mm,n])=1);
 if nn::integer and nn > mm then
 printf("Found [%d, %d, %d, %d, %d]n",jj,kk,ll,mm,nn)
 fi
 od od od od:

$2,3,7,43,1807 $ - the first 5 terms of Sylvester's sequence - also works. In this sequence each term is the product of the previous terms plus $1$.

So it looks like the solution is not unique.

(Just saw that Robert Israel already made this observation).

Searching through brute force gives a solution $2, 3, 11, 23, 31 $

Assume $J < K < L < M < N$ and
also note that the least number $J$ can only be $2$ or $3$

In Python $3.x$, you can check by running this code

for j in range(2, 4):
 for k in range(j+1, 100):
 for l in range(k+1, 100):
 for m in range(l+1, 100):
 for n in range(m+1, 100):
 if k*l*m*n + j*l*m*n + j*k*m*n + j*k*l*n + j*k*l*m + 1 == j*k*l*m*n:
 print(j, k , l , m , n)

Let's approach the problem one variable at a time. Without loss of generality, assume that $J < K < L < M < N$.

What is J?

If $J = 1$, then we would have $frac1K + frac1L + frac1M + frac1N + frac1KLMN = 0$, which is clearly impossible. So $J ne 1$.

If $J ≥ 4$, then the greatest the LHS could possibly be is $frac14 + frac15 + frac16 + frac17 + frac18 + frac14⋅5⋅6⋅7⋅8 = frac11891344 < 1$. And increasing any variable simply makes a smaller fraction. It will always be less than 1. So, any solution with $J ≥ 4$ is ruled out.

OTOH, $J = 3$ produces an upper bound of $frac13 + frac14 + frac15 + frac16 + frac17 + frac13⋅4⋅5⋅6⋅7 = frac551504 > 1$, which is OK.

So, $J in lbrace 2, 3 rbrace$.

What is K?

Since there are only two possibilities for $J$, let's plug in each of them.

• If $J = 2$, then $frac1K + frac1L + frac1M + frac1N + frac12KLMN = frac12$. As before, the LHS is maximized by taking all the variables to be consecutive integers.
  
  
  • If $K = 6$, then we have $frac16 + frac17 + frac18 + frac19 + frac12⋅6⋅7⋅8⋅9 = frac33016048 > frac12$, which is fine.
  
  
  • But if $K = 7$, we have $frac17 + frac18 + frac19 + frac110 + frac12⋅7⋅8⋅9⋅10 = frac482910080 < frac12$, which is too low. So $K ≤ 6$.
  
  
  • Recalling that $K > J$, this means $K in lbrace 3, 4, 5, 6 rbrace$.
  
  


• If $J = 3$, then $frac1K + frac1L + frac1M + frac1N + frac13KLMN = frac23$.
  
  
  • If $K = 4$, then the upper bound on the LHS is $frac14 + frac15 + frac16 + frac17 + frac13⋅4⋅5⋅6⋅7 = frac383504 > frac23$, which is OK.
  
  
  • But if $K = 5$, then we have $frac15 + frac16 + frac17 + frac18 + frac13⋅5⋅6⋅7⋅8 = frac457720 < frac23$, which is too low.
  
  
  • So $K = 4$ is the only possibility.
  
  

Taking the union of the cases, we have $K in lbrace 3, 4, 5, 6 rbrace$.

What is L?

From the previous section, we have 5 possibilities for $(J, K)$:

• 
  $J = 2$, $K = 3$. Then $frac1L + frac1M + frac1N + frac16LMN = frac16$, and $4 ≤ L ≤ 17$.


• 
  $J = 2$, $K = 4$. Then $frac1L + frac1M + frac1N + frac18LMN = frac14$, and $5 ≤ L ≤ 11$.


• 
  $J = 2$, $K = 5$. Then $frac1L + frac1M + frac1N + frac110LMN = frac310$, and $6 ≤ L ≤ 9$.


• 
  $J = 2$, $K = 6$. Then $frac1L + frac1M + frac1N + frac112LMN = frac13$, and $7 ≤ L ≤ 8$.


• 
  $J = 3$, $K = 4$. Then $frac1L + frac1M + frac1N + frac112LMN = frac512$, and $5 ≤ L ≤ 6$.

Taking the union of these gives $4 ≤ L ≤ 17$.

What is M?

If we take the minimum values for the other variables: $J = 2$, $K = 3$, and $L = 4$, then $frac12 + frac13 + frac14 + frac1M + frac1N + frac124MN = 1$, or $frac1M + frac1N + frac124MN = frac-112$. That negative number on the right means that the approach used to find an upper bound for J, K, and L won't work for M. So, let's just skip it and come back to it later.

What is N?

If we have values for the other 4 variables, then we can solve for N directly.

$$frac1J + frac1K + frac1L + frac1M + frac1N + frac1JKLMN = 1$$

$$frac1N(1 + frac1JKLM) = 1 - (frac1J + frac1K + frac1L + frac1M)$$

$$frac1N = frac1 - (frac1J + frac1K + frac1L + frac1M)1 + frac1JKLM$$

$$N = frac1 + frac1JKLM1 - (frac1J + frac1K + frac1L + frac1M)$$

$$N = fracJKLM + 1JKLM - (KLM + JLM + JKM + JKL)$$

All we have to do is confirm that this number is an integer, and that it is greater than $M$.

Brute force

A slight modification of ab123's Python script to use my tighter bounds for J, K, and L; and formula for N.

from fractions import Fraction

MAX_M = 1000000

for J in range(2, 4):
 for K in range(J + 1, 7):
 for L in range(K + 1, 18):
 for M in range(L + 1, MAX_M + 1):
 N1 = J*K*L*M + 1
 N2 = J*K*L*M - (K*L*M + J*L*M + J*K*M + J*K*L)
 if N2 != 0:
 N = Fraction(N1, N2)
 if N.denominator == 1 and N > M:
 print(J, K, L, M, N)

This gives three solutions:

• (2, 3, 7, 43, 1807)


• (2, 3, 7, 47, 395)


• (2, 3, 11, 23, 31)

Perhaps other solutions exist with $M > 10^6$. Or someone can prove that they don't.