Can we add a ordinal bigger than every other ordinal and end up with a transitive model? Given some (set sized) transitive model of ZFC, $M$ we can construct Hyper-$M$ as follows. We construct an ultrafilter $U$ on $Ord_M$ such that $(\exists \alpha.S = \{ \beta : \beta > \alpha \}) \implies S \in U$. Hyper-$M$ is then $M^{|Ord_M|}/U$ (by analogy with the name Hyperreals).
Note that Hyper-$M$ is an elementary extension of $M$, and moreover is contains an ordinal greater than every ordinal in $M$, namely $H = (0,1,2,\dots,\beta,\dots)/U$. This is because for any ordinal $\alpha = (\alpha, \alpha, \dots, \alpha, \dots)/U$, the set of indices $H_i > \alpha_i$ is equal to $\{ \beta : \beta > \alpha \} \in U$.
My question:

Is Hyper-$M$ isomorphic to some transitive model of ZFC? If not, how about an elementarily equivalent transitive model?

 A: Building off the comments to the question, I think the following is the answer:
First, a quick fact:


*

*The following are equivalent for a set model $M$ of ZFC:


*

*$M$ is isomorphic to a transitive model of ZFC.

*$M$ is well-founded.

*$Ord^M$ is well-ordered with respect to $\in^M$.

*For each $\alpha\in Ord^M$, $\{m\in M: m\in^M\alpha\}$ is well-ordered with respect to $\in^M$.
Clearly $(1)$ implies $(2)$, and from $(2)$ and the Mostowski collapse we get $(1)$. The equivalence between $(3)$ and $(4)$ is because any linear order all of whose proper initial segments are well-ordered is itself a well-order. $(2)$ clearly implies $(3)$, and from $(3)$ and thinking about rank in the cumulative hierarchy we get $(2)$.
With this in mind, you are ultimately asking:

Suppose we have a well-founded model $M$ of ZFC and $\mathcal{U}$ is an ultrafilter on $Ord^M$ such that every "up-set" is in $\mathcal{U}$. Need $\prod_{Ord^M} M/\mathcal{U}$ be well-founded?

This is ultimately addressed by the notion of completeness of an ultrafilter. An ultrafilter $\mathcal{W}$ on a set $I$ is $\kappa$-complete if for every $S\subseteq\mathcal{W}$ with $\vert S\vert<\kappa$ we have $\bigcap S\in\mathcal{W}$, that is, if $\mathcal{W}$ is closed under size-less-than-$\kappa$ intersections. Every ultrafilter is $\omega$-complete, and the first interesting situation occurs when we consider $\omega_1$-completeness (also called countable completeness). Here's the key observation:


*

*Suppose $\mathcal{L}$ is an infinite linear order, $I$ is some index set, and $\mathcal{W}$ is an ultrafilter on $I$ which is not $\omega_1$-closed. Then - even if $\mathcal{L}$ was well-ordered! - $\prod_I\mathcal{L}/\mathcal{U}$ is ill-founded.


Conversely, we have:


*

*If $\mathcal{L}$ is a well-ordering and $\mathcal{U}$ is an $\omega_1$-closed ultrafilter on an arbitrary index set $I$, then $\prod_I\mathcal{L}/\mathcal{U}$ is well-ordered.


In combination with the "quick fact" above (hint: take $\mathcal{L}$ to be $Ord^M$) this tells us:

If $M$ is a transitive model of ZFC, then $\prod_IM/\mathcal{U}$ is isomorphic to a transitive model of ZFC iff $\mathcal{U}$ is $\omega_1$-closed. 

By a standard Zorn's lemma argument, we can cook up non-$\omega_1$-closed appropriate ultrafilters, so the answer to your question is no. Indeed, well-foundedness is hard to preserve: the existence of an $\omega_1$-complete (nonprincipal) ultrafilter on any set is equivalent to the existence of a measurable cardinal, which is a fairly powerful large cardinal axiom (the biggest of the smalls and the smallest of the bigs, really).
A: Noah did an excellent job answering the body of your question. Just for fun I will ignore the body of your question and answer the question in your title instead in a slightly different, but certainly related, way:

Can we add a[n] ordinal bigger than every other ordinal and end up with a transitive model [via an ultrapower construction]?

The answer is again 'yes'. Under mild assumptions (much weaker than the existence of a measurable cardinal) there is a structure (a so-called mouse) $0^{\#} = (L_\alpha; \in, F)$, where $L_{\alpha}$ is a countable initial segment of the constructible universe and $F$ is a (not $\omega_1$-complete) ultrafilter on some $\kappa$ such that 
$$(L_{\alpha}; \in) \models \mathrm{ZFC}^{-} \text{ and } \kappa \text{ is the largest cardinal}
$$
with the property that $\mathrm{Ult}(L_{\alpha}; F) = (X; E)$ is well-founded. Here


*

*$X$ consists of all $\sim$-equivalence classes $[f]$ for $f \colon \kappa \to L_{\alpha}$, $f \in L_{\alpha}$ and
$$
f \sim g : \iff \{ \xi < \kappa \mid f(\xi) = g(\xi) \} \in F,
$$

*$E$ is a relation on $X \times X$, intended to interpret $\dot{\in}$, given by
$$
[f] E [g] : \iff \{ \xi < \kappa \mid f(\xi) \in g(\xi) \} \in F.
$$


Since $(X; E)$ is also extensional, it will be isomorphic to some transitive structure $(X; E) \cong (Y; \in)$ and in fact $Y = L_{\beta}$ for some $\beta > \alpha$. In particular $L_{\alpha} \cap \mathrm{Ord} = \alpha < \beta = L_{\beta} \cap \mathrm{Ord}$.
