The answers in the two cases are clearly different, so a formula for one of the problems cannot possibly work, without change, to solve the other. A reasonable way to approach the question is to solve each of the problems, in the concrete case you mentioned. We use approaches that readily generalize.
Different toys: Call the toys $1$, $2$, and so on up to $15$ (numbers make great toys). There are $3$ ways to decide who gets toy $1$. For each of these ways, there are $3$ ways to decide who gets toy $2$, for a total so far of $3\times 3$. For each way of making these two decisions, there are $3$ ways to decide who gets toy $3$, and so on, for a total of $3^{15}$ ways.
We could approach the problem through multinomial coefficients. The number of ways to choose $t_A$ toys to give to kid $A$, and $t_B$ toys to give to kid $B$, and $t_C$ to give to kid $C$, where $t_A+t_B+t_B=15$, is the multinomial coefficient $\binom{15}{t_A,t_B,t_C}$. To get the total number of ways, sum over all $(t_A,t_B,t_C)$. We can use the multinomial theorem to conclude that the sum is $3^{15}$. But we already had a simpler argument for $3^{15}$.
Identical toys: A standard approach is to count the number of ways to distribute $18$ toys among the $3$ kids, at least one toy to each kid. Then we make each kid give back a toy. Line up the $18$ toys. They determine $17$ inter-toy gaps. Put a marker into $2$ of these gaps. Give kid $A$ all the toys up to the first marker, kid $B$ the toys from the first marker to the second, and kid $C$ the rest. There are $\binom{17}{2}$ ways to choose where the markers will go, and hence $\binom{17}{2}$ ways to distribute $15$ toys among $3$ kids, where some kid(s) may get nothing.
The argument perhaps is not *multi*nomial, since only the special binomial case of the multinomial is being used. But it certainly belongs to the same family of ideas. In both cases, "nomial" ideas can be used.