Let $C_k$ be the set of partitions of $n$ containing $k$.

Following @MaxAlekseyev's point we have,
$$P(n-1)+P(n-2)-P(n)=|C_1|+|C_2|-P(n),$$
$$=|C_1\cap C_2|+|C_1\cup C_2|-P(n)$$
This is the # of partitions containing both 1 and 2, minus the # of partitions containing neither 1 nor 2.

This number is nonnegative since to any partition containing neither 1 nor 2, writing it in nonincreasing order as $t_1,\dots, t_k$, we can associate the partition $t_1,\dots,t_k-3,2,1$ which does contain 1 and 2 (and this map is one-to-one).

Thus for $n\ge 4$ certainly, $P(n-1)+P(n-2)-P(n)$ is exactly counting 
>how many partitions of $n$ contain both 1 and 2, and are not of the form $t_1,\dots,t_{k-1},t_k-3,2,1$ where $t_1\ge\dots\ge t_k\ge 3$.

For $n=5$, this includes only one: $2+1+1+1$.