$\newcommand{\Ord}{\text{Ord}}
\newcommand{\ZFC}{\text{ZFC}}$

Here is one way to formalize your concept a little more tightly,
which provides the answers to your questions. For any large
cardinal property $P$, let's take the phrase "*$\Ord$ is $P$*" to
be the theory asserting $\sigma$, for any sentence $\sigma$ that
ZFC proves is true in $V_\delta$ under the assumption that
$\delta$ has property $P$ in $V$.

With this formalization, "$\Ord$ is $P$" asserts that the universe is just like we would expect, if we were living inside $V_\delta$ for an actual $P$ cardinal $\delta$. For example, "$\Ord$ is measurable" implies that there are a proper
class of weakly compact cardinals, since this is true in
$V_\delta$, whenever $\delta$ is measurable, and "$\Ord$ is
supercompact" implies that there are many partially supercompact
cardinals, with nice limit properties. For example, they would
form a stationary class in the sense that every definable class
club would contain one of them. This notion seems to capture what
one would want to mean by saying $\Ord$ has property $P$ as a
purely first-order theory about sets.

With this idea, the point I would like to make is that assuming
$\Ord$ is $P$ is essentially equivalent to assuming that what you
have is $V_\delta$, where $\delta$ has property $P$ in a larger
universe.

**Theorem.** For any large cardinal property $P$, a model of set
theory $M$ satisfies "$\Ord$ is $P$" if and only if $M\prec
V_\delta^N$ for some taller model of set theory $N$ with a
cardinal $\delta$ having property $P$ in $N$.

Proof. The backward direction is immediate, since $\delta$ having
property $P$ implies that $V_\delta$ satisfies every assertion of
$\Ord$ is $P$. For the forward direction, suppose $M$ satisfies
$\Ord$ is $P$. Let $T$ be the theory consisting of $\ZFC$, plus the
assertion "$\delta$ is $P$", using a new constant symbol $\delta$,
plus the assertions $\varphi^{V_\delta}$, for any $\varphi$ in the
elementary diagram of $M$, using constants for elements of $M$.
This theory is finitely consistent, since otherwise there would be
finitely many assertions in it that are contradictory, and so
there would be a statement $\varphi$ true in $M$ that provably
could not hold in $V_\delta$ for any cardinal $\delta$ with
property $P$. But that would contradict our assumption that
$M\models\Ord$ is $P$.

If $N$ is any model of the theory, then $\delta$ has property $P$
in $N$, and we get $M\prec V_\delta^N$, because $V_\delta^N$
satisfies the elementary diagram of $M$. Another way to say this
is that there is an elementary embedding $j:M\to V_\delta^N$,
mapping every element of $M$ to the interpretation of its constant
in $N$. QED

Thus, if one is inclined to assume $\Ord$ is $P$, then why not go ahead and make the full move to a model with an actual $P$ cardinal $\delta$, such that our old
world looks exactly like $V_\delta$ in this new world. In particular, under this terminology, the theory $\ZFC+\Ord$ is $P$ is equiconsistent with $\ZFC+\exists \delta$ with property $P$.

**Corollary.** The following theories are equiconsistent:

- $\ZFC+\Ord$ is $P$.
- $\ZFC+\exists \kappa$ with property $P$.

Lastly, I would like to point out that there is some variance in
the literature about what "$\Ord$ is $P$" should mean. For
example, one often finds the phrase "$\Ord$ is Mahlo" to mean only
the weaker assertion, that every definable closed unbounded class
of cardinals contains a regular cardinal. This is what one finds, for example,
at Cantor's Attic. But this is strictly weaker
in consistency strength than ZFC+$\exists\kappa$ Mahlo, since this latter theory implies
the consistency of the former, as it is true in $V_\kappa$
whenever $\kappa$ is Mahlo.