I don't know the answer to the original question, but I don't think I'd be alone in arguing that the condition "there is an adjunction terminal among those inducing $T$" doesn't "feel" like "the right condition". Let me argue that you can massage this into a "nicer" condition which still expresses the universal property of the EM construction in terms of adjunctions:
Claim: Let $\mathcal K$ be a 2-category. Then $\mathcal K$ admits Eilenberg-Moore objects if and only if the forgetful functor $Adj(\mathcal K) \to Mnd(\mathcal K)$ has a right adjoint.
Here $Mnd(\mathcal K)$ is defined as usual, and $Adj(\mathcal K)$ is the 2-category of adjunctions in $\mathcal K$, where a 2-cell is a 2-cell between the right adjoints.
Proof: As you say, Street proves one direction, so we prove the other. We have forgetful 2-functors
$$ Mnd(\mathcal K) \leftarrow Adj(\mathcal K) \xleftarrow{const} \mathcal K $$
By hypothesis, the leftmost of these has a right adjoint. We wish to show that the composite (sending $K \in \mathcal K$ to the identity monad on $K$) has a right adjoint (this characterization of EM objects is of course also due to Street in the same paper). It will suffice to show that the rightmost 2-functor has a right adjoint. And it does -- the right adjoint sends $(L : C {}^\to_\leftarrow D : R) \mapsto D$.
Notes:
If you think about the proof above, you'll see it also shows that this is true for a particular universal adjunction / EM object, even if these constructions aren't assumed to exist in general.
(The funny thing here is that the diagram above doesn't seem to be induced in an obvious way by somehow "homming" some diagram into $\mathcal K$. At least $Mnd(\mathcal K)$ is the lax Gray internal hom $[[ Mnd, \mathcal K]]$ -- or equally the lax functor category $Fun^{lax}(1, \mathcal K)$ -- but I don't know how to construct $Adj(\mathcal K)$ except "by hand". As a result, the verifications in the above also need to be done "by hand". But it works out!)
The condition that there be a terminal adjunctions inducing given monads $T$ says that the fiber of $Adj(\mathcal K) \to Mnd(\mathcal K)$ over $T$ has a terminal object. The condition that the right adjoint to $Adj(\mathcal K) \to Mnd(\mathcal K)$ exist is stronger in one sense. But it's also weaker, unless we have some way of knowing that this right adjoint should be fully faithful -- a question which Street's proof of the other direction of the Claim should shed some light on.