Here's an elaboration of my comment above in the pairwise coprime case, in an even more special case (the fact that in a finite solvable group we can't have three elements of pairwise coprime orders whose product is the identity is, I believe, due to P. Hall). If one insists, as in the question, on the minimality of the order, then if $a,b,c$ are distinct primes, any finite group $G$ of minimal order subject to containing elements $x,y,z$ of respective orders $a,b,c$ such that $xyz = 1_{G}$ (equivalently, $xy = z^{-1}$), is necessarily a finite simple group.

For let $N$ be a proper non-identity normal subgroup of $G.$ Then if none of $x,y,z$ is in $N,$ the minimality of $G$ is contradicted, as $xN yN zN = N$ in $G/N.$ If two of $x,y,z$ are in $N,$ then so is the third, and the minimality of $G$ is contradicted again.If (say) $x \in N,$ but neither $y$ nor $z$ lies in $N,$ we have a contradiction since $yNzN = N$ in $G/N$, but $yN$ and $zN$ have coprime orders. Hence there is no such normal subgroup $N$ and $G$ is simple.

Later edit: Here is a general argument which may be relevant, and admits various generalizations: Let $a>5$ be an odd prime, $b>1$ be an odd divisor of $a-1$ and $c>1$ be an odd divisor of $a+1$ (note that $a$ is then neither a Fermat nor a Mersenne prime). Then $G = {\rm PSL}(2,a)$ (which is isomorphic to a subgroup of $S_{a+1}) $ contains three elements $x,y,z$ of respective orders $a,b,c$ with $xyz = 1.$
This is because if $u$ is an element of order $a,v$ an element of order $b$ and $w$ an element of order $c$ in $G,$ then each non-trivial irreducible character $\chi$ of $G$ vanishes at one of $u,v,w.$ Hence $\sum_{\chi \in {\rm Irr}(G)} \frac{\chi(u) \chi(v) \chi(w)}{\chi(1)}> 0.$ By a standard character-theoretic formula,this means that we may choose conjugates $x,y,z$ of $u,v,w$ respectively such that $xy = w^{-1},$ or $xyw = 1.$
It is also easy to check that $G = \langle x,y \rangle$ for any such $x,y$, since $\langle x,y \rangle$ must have more that one Sylow $a$-subgroup, so has $a+1$ Sylow $a$-subgroups and hence is all of $G.$

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