The idea is to build in ZFC using replacement, a set REPLACEMENT(x ∈ A: TERM(x))
from a set and a term in the same way the set {x ∈ A: FORMULA(x)} is built
using specification from a set and a formula.


Given a set A and a TERM(x) with a free variable x, can be proved, using 
the axiom scheme of replacement, the existence of a set named


REPLACEMENT(x ∈ A:TERM(x)) with the following property


For all set z, z ∈ REPLACEMENT(x ∈ A:TERM(x)) <==> 
There is one x ∈ A such that z = TERM(x) 



Possible no standard use of replacement axiom
If you're just asking whether this can be proved, the answer is yes (assuming that "For all SET(z)" means "For all sets z"). More importantly, this can be proved also for extensions by definitions of ZFC. That's important, because the official vocabulary of ZFC has no function symbols, so the only terms are variables; you need to pass to extensions by definitions to get non-trivial terms.


I don't like your notation though; it would be less confusing if the elements of the resulting set were to the left of the colon, as in the usual set-builder notation. 


A more general notation, $\{t(x):x\in A:\phi(x)\}$, where $t(x)$ is a term and $\phi(x)$ is a formula, has been used. I don't remember where I first saw it; Yuri Gurevich, Saharon Shelah, and I used it in our paper "Choiceless polynomial time", but it had already been used earlier by others.


