There was this dog named Rita

It's not too hard.

First, show that $a, b, c$ can't all be prime. Suppose not, then we have $d ( ab + bc + ca) = abc$ . Taking mod $a$ , we get $dbc \equiv 0 \pmod{a}$ , and so $a \mid d$ . By similarity, $abc \mid d$ . But $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{abcd^*}$ is absurd.

WLOG, let $c$ be the non-prime. Clearly, $c \neq 1$ , so $c$ is composite. Let's denote it by $x$ , and $a, b, d$ are all prime. We have

$d ( ab + bx + xa) = abx$

Since LHS is divisible by $d$ , hence $d \mid x$ . Since RHS is divisible by $a$ , hence $a \mid dbx \Rightarrow a \mid x$ . Since RHS is divisible by $b$ , hence $b \mid dxa \Rightarrow b \mid x$ . Since $a,b,d$ are distinct primes, thus $abd \mid x$ . Let $x = abdx^*$ .

This gives us $d(ab + babdx^* + aabdx^*) = ababdx^*$ . Factoring out $abd$ , we get $1 + dbx^* + dax^* = abx^*$ . Taking modulus $x^*$ , we conclude that $x^* = 1$ , so $x = abd$ .

Now, we have $1 + db + da = ab$ , or that $(a-d)(b-d) = d^2 + 1$ .
Case 1: If $d = 2$ , then we have $(a-2)(b-2) = 5$ , which gives us the solution $\{ a, b \} = \{3, 7 \}$ . Check that $2, 3, 7$ are all primes.
Case 2: If $d \neq 2$ , then since $d$ is prime, so $d$ is odd. Clearly, $a$ and $b$ have to be primes larger than $d$ , so they are also odd. Thus $a-d$ and $b-d$ are both even, so $4 \mid (a-d)(b-d)$ . But we know that $4 \not \mid d^2 + 1$ , hence there are no solutions in this case.