We can find a bivariate generating function that is exponential in $m$ (the number of parts), and ordinary in $n$. Other variants are harder questions, so I suspect this is what the exercise intended. We use $z$ to mark size ($n$), and $u$ to mark the number of parts ($m$).
Fix $m$, the number of parts. For any positive integer $k$, if we denote by $c_k(n,m)$ the number of compositions of $n$ into $m$ non-unique parts all equal to $k$ (thus $c_k(n,m) = 1$ when $n = mk$ and $m \neq 1$, and $0$ otherwise), then the generating function
$$C_k(z, u) = \sum_{m,n \ge 0} c_k(n,m) z^n \frac{u^m}{m!} = \sum_{m \ge 0, m \neq 1} z^{mk} \frac{u^m}{m!} = e^{uz^k} - uz^k$$
When we allow compositions using two integers $k$ and $l$, the corresponding number of compositions $c_{k, l}(n, m)$ forms a binomial convolution, corresponding to which of the $m$ parts are chosen to be equal to $k$, and which equal to $l$:
$$c_{k, l}(n, m) = \sum_{a \ge 0} \binom{m}{a} c_k(ak, a) c_l(n - ak, m - a)$$
which corresponds to a product of generating functions:
$$C_{k,l}(z, u) = \sum_{m,n \ge 0} c_{k,l}(n,m) z^n \frac{u^m}{m!} = (\exp(uz^k) - uz^k)\cdot (\exp(uz^l) - uz^l)$$
Thus the class of all restricted compositions has generating function
$$C(z, u) = \prod_{k=1}^{\infty}\left(\exp(uz^k) - uz^k\right)$$
Further simplification doesn't seem to be possible.
