Too much.

Multiply three relations to get $abc \mid abc(abc+a+b+c)+ab+ac+bc+1$ . Thus $abc \mid ab+ac+bc+1$ and the expression $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{abc}$ must be a positive integer. If $a \ge 4$ , then $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{abc} \le \frac{3}{4}+\frac{1}{64}<1$ Thus $a \le 3$ . For $a=1$ , using the same reasoning for bounding $a$ , one can conclude that $b \le 2$ . Similarly for $a=2$ we get $2 \le b \le 4$ and for $a=3$ we get $b=3$ .

For $(a, b)=(1, 1)$ we get $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{abc}=2+\frac{2}{c}$ . It gives two solutions $(a, b, c)=(1, 1, 1), (1, 1, 2)$ . The first one negates $c \neq ab$ and the second negates $c \neq a+b$ .

If $(a, b)=(1, 2)$ we get $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{abc}=1+\frac{c+3}{2c}$ . Thus $c=1$ or $c=2$ . Hence $2c \mid 2(c+3)-2c=6$ and $c \mid 3$ . Thus $c=3$ . It gives the solution $(a, b, c)=(1, 2, 3)$ . This solution negates the condition $c \neq a+b$ .

For $(a, b)=(2, 2)$ we get $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{abc}=1+\frac{5}{4c}$ , which is never an integer.

If $(a, b)=(2, 3)$ we get $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{abc}=\frac{1}{2}+\frac{1}{3}+\frac{1}{c}+\frac{1}{6c}=\frac{5c+7}{6c}$ . Hence $6c \mid 6(5c+7)-5(6c)=42$ and $c \mid 7$ . Thus $c=7$ . It gives the solution $(a, b, c)=(2, 3, {\color{#D61F06}7})$ .

For $(a, b)=(2, 4)$ we get $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{abc}=\frac{1}{2}+\frac{1}{4}+\frac{1}{c}+\frac{1}{8c}=\frac{6c+9}{8c}$ , which is an odd number divided by an even number and never an integer.

Finally for $(a, b)=(3, 3)$ we get $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{abc}=\frac{1}{3}+\frac{1}{3}+\frac{1}{c}+\frac{1}{9c}=\frac{6c+10}{9c}$ , which is never an integer since numerator is not divisible by $3$ .