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 $frac{1}{K} + frac{1}{L} + frac{1}{M} + frac{1}{N} + frac{1}{KLMN} = 0$, which is clearly impossible. So $J ne 1$.

If $J ≥ 4$, then the greatest the LHS could possibly be is $frac{1}{4} + frac{1}{5} + frac{1}{6} + frac{1}{7} + frac{1}{8} + frac{1}{4⋅5⋅6⋅7⋅8} = frac{1189}{1344} < 1$. 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 $frac{1}{3} + frac{1}{4} + frac{1}{5} + frac{1}{6} + frac{1}{7} + frac{1}{3⋅4⋅5⋅6⋅7} = frac{551}{504} > 1$, 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 $frac{1}{K} + frac{1}{L} + frac{1}{M} + frac{1}{N} + frac{1}{2KLMN} = frac{1}{2}$. As before, the LHS is maximized by taking all the variables to be consecutive integers.
  
  
  
  
  • If $K = 6$, then we have $frac{1}{6} + frac{1}{7} + frac{1}{8} + frac{1}{9} + frac{1}{2⋅6⋅7⋅8⋅9} = frac{3301}{6048} > frac{1}{2}$, which is fine.
  
  
  • But if $K = 7$, we have $frac{1}{7} + frac{1}{8} + frac{1}{9} + frac{1}{10} + frac{1}{2⋅7⋅8⋅9⋅10} = frac{4829}{10080} < frac{1}{2}$, 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 $frac{1}{K} + frac{1}{L} + frac{1}{M} + frac{1}{N} + frac{1}{3KLMN} = frac{2}{3}$.
  
  
  
  
  • If $K = 4$, then the upper bound on the LHS is $frac{1}{4} + frac{1}{5} + frac{1}{6} + frac{1}{7} + frac{1}{3⋅4⋅5⋅6⋅7} = frac{383}{504} > frac{2}{3}$, which is OK.
  
  
  • But if $K = 5$, then we have $frac{1}{5} + frac{1}{6} + frac{1}{7} + frac{1}{8} + frac{1}{3⋅5⋅6⋅7⋅8} = frac{457}{720} < frac{2}{3}$, 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 $frac{1}{L} + frac{1}{M} + frac{1}{N} + frac{1}{6LMN} = frac{1}{6}$, and $4 ≤ L ≤ 17$.


• 
  $J = 2$, $K = 4$. Then $frac{1}{L} + frac{1}{M} + frac{1}{N} + frac{1}{8LMN} = frac{1}{4}$, and $5 ≤ L ≤ 11$.


• 
  $J = 2$, $K = 5$. Then $frac{1}{L} + frac{1}{M} + frac{1}{N} + frac{1}{10LMN} = frac{3}{10}$, and $6 ≤ L ≤ 9$.


• 
  $J = 2$, $K = 6$. Then $frac{1}{L} + frac{1}{M} + frac{1}{N} + frac{1}{12LMN} = frac{1}{3}$, and $7 ≤ L ≤ 8$.


• 
  $J = 3$, $K = 4$. Then $frac{1}{L} + frac{1}{M} + frac{1}{N} + frac{1}{12LMN} = frac{5}{12}$, 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 $frac{1}{2} + frac{1}{3} + frac{1}{4} + frac{1}{M} + frac{1}{N} + frac{1}{24MN} = 1$, or $frac{1}{M} + frac{1}{N} + frac{1}{24MN} = frac{-1}{12}$. 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.

$$frac{1}{J} + frac{1}{K} + frac{1}{L} + frac{1}{M} + frac{1}{N} + frac{1}{JKLMN} = 1$$

$$frac{1}{N}(1 + frac{1}{JKLM}) = 1 - (frac{1}{J} + frac{1}{K} + frac{1}{L} + frac{1}{M})$$

$$frac{1}{N} = frac{1 - (frac{1}{J} + frac{1}{K} + frac{1}{L} + frac{1}{M})}{1 + frac{1}{JKLM}}$$

$$N = frac{1 + frac{1}{JKLM}}{1 - (frac{1}{J} + frac{1}{K} + frac{1}{L} + frac{1}{M})}$$

$$N = frac{JKLM + 1}{JKLM - (KLM + JLM + JKM + JKL)}$$

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.