$P+Q-PQ$ is a projection if and only if $PQ=QP$.

If $P+Q-PQ$ is a projection, then \begin{align} P+Q-PQ &= (P+Q-PQ)^*\\ &= P^* + Q^* - (PQ)^*\\ &= P^* + Q^* - Q^*P^*\\ &= P + Q - QP, \end{align} from which it follows that $PQ=QP$.

To complement Math1000's answer, the path of working with the equality $(P+Q-PQ)^2=P+Q-PQ$ cannot lead to a proof (i.e., the argument does not work for idempotents alone). Let $$ P=\begin{bmatrix}1&1\\0&0\end{bmatrix},\ \ Q=\begin{bmatrix}1&0\\0&0\end{bmatrix}. $$ Then $PQ=Q$, $QP=P$, and $$P+Q-PQ=P=(P+Q-PQ)^2.$$