The answer rather depends on what you mean by an "algebra". For instance, universal algebraists would certainly call groups a kind of algebra, and there's no operad for which the algebras are groups.

For operads of sets (i.e. operads where the operations of each arity form a set rather than, say, a vector space or topological space), it's well-understood which algebraic theories can be obtained. I'll split the answer into two, according to whether or not you wish your operads to come equipped with a symmetric group action.

The algebraic theories that arise from non-symmetric operads are exactly the so-called *strongly regular* theories. These are the finitary algebraic theories that can be presented using only equations in which the same variables appear in the same order, and only once each, on each side. For example, the associativity equation $(xy)z = x(yz)$ is fine, but the inverse equation $x x^{-1} = 1$ is not (for two reasons: $x$ appears twice on the left-hand side, and $x$ appears on the left-hand side but not the right-hand side).

The algebraic theories that arise from symmetric operads can be characterized similarly, just dropping the "in the same order" requirement.

For operads of vector spaces, I guess there are similar theorems. E.g. the Jacobi identity $[x,[y,z]] + [y,[z,x]] + [z,[x,y]] = 0$ is OK because although you've got variables appearing several times on one side of the equation, each variable appears exactly once in each summand. As far as I know, though, no one's actually *proved* a theorem like this. Someone should!

(Note that although the equation $x x^{-1} = 1$ doesn't satisfy the requirements above, it doesn't follow from this that there is no operad whose algebras are exactly groups. Conceivably there is some other, sneaky presentation of the theory of groups that uses only equations satisfying the requirements. In fact, there isn't — it can be proved that there really is no operad for groups. But that requires a different theorem.)