Equating
#{(a*b = a/b + e_1^2), (a/b = a - b + e_2^2), (a - b = a + b + e_3^2):}#
and solving for #a,b,e_1#
#{(a = -(2 e_2^2 e_3^2 + e_3^4)/(2 (2 + e_3^2))), (b = -e_3^2/2),
(e_1 = -1/2 sqrt[e_3^2-2] sqrt[2 e_2^2 + e_3^2]):}#
but #e_1# must be real so #e_3^2 > 2# and we choose #e_3 = sqrt3#
and now solving
#{(a*b = a/b + e_1^2), (a/b = a - b + e_2^2), (a - b = a + b + 3):}#
for #a,b,e_1#
#{(a = -3/10(3 + 2 e_2^2)), (b = -3/2),
(e_1 = -1/2 sqrt[3 + 2 e_2^2]):}#
putting now #e_2 = 2# we have
#{(a = -33/10), (b = -3/2), (e_1 = sqrt[11]/2):}#
and finally one of the infinite solutions
#a=-33/10, b = -3/2#