There is a confusion here. Recursion goes up, based on a well-founded relation, it does not go down. When you define something by recursion it is a step-by-step definition. From assuming you could define it so far, you define it one step further.
Your definition seems to be co-well-founded, you postulate the property of a set and require it implies the property for its elements.
In $\sf ZF$ the $\in$ relation is well-founded, so this is not quite possible. But we can still get some sufficiently nice properties which will hold for the elements. For example being hereditarily ordinal definable, or constructible. Or being an ordinal. Or just being a set.
But in none of these cases you start by postulating the property for a set, and then require it implies the property for the elements of the set. Instead you start by construction some class, and taking its hereditary part, or you define something from bottom to top, and then you end up with a class satisfying your property.
When you tack on the requirement that the property implies that the set is infinite, then you quickly run into a contradiction. The bottom-up constructions start with the empty set—for a reason, too: it's the basis of the set theoretic universe—and go upwards. So at the root of every set, you will find the empty set. And so in every non-empty transitive set, you will find the empty set as an element.
If you require that $P$ is hereditary, then it implies that if $x$ has it, the transitive closure of $x$ has it. But if $x$ is non-empty, $\varnothing$ is an element of that transitive closure. So infinitude cannot be part of the definition of $P$.