The effect of the left adjoint $F$ on morphisms is already determined by the universal property of the units of the adjunction: given a map $f: x \to y$, consider the diagram
$$\matrix{
x & \mathop{\longrightarrow}\limits^{\eta_x} & GFx \cr
{\scriptstyle f} \big\downarrow {\ } & & \cr
y & \mathop{\longrightarrow}\limits_{\eta_y} & GFy \cr}
$$
and let $Ff: Fx\to Fy$ be the unique map that corresponds to the composite $\eta_y\circ f: x\to GFy$.
In your example, if $(a,h:Ta\to a)$ is a $T$-algebra, a map $g: x\to a$ corresponds to $h\circ Tg: (Tx,\mu_x)\to (a,h)$ and consequently, for $f: x\to y$ we obtain:
$Ff = \mu_y\circ T(\eta_y\circ f) = \mu_y\circ T\eta_y \circ Tf = Tf$.
Hint 1: we already have an arrow $C \to TC$ since $T$ is a monad. Since $\mathcal{C}$ is a poset category, saying that $TC = C$ is just the same as saying that we have an arrow $TC \to C$.
Hint 2: since $\mathcal{C}$ is a poset category, all diagrams commute automatically, simplifying the definition of the Eilenberg-Moore category considerably.
The below was written before I noticed you said poset category. Feel free to ignore this.
I don't think this is true as stated. The double powerset monad on $\mathcal{Set}$ with $TX = P(P(X))$ has no $T$-closed objects. Even if we loosen $T$-closed to mean $TX \cong X$ (which may be what you meant to start with), the double powerset of $X$ has much larger cardinality than $X$. So the full subcategory of $\mathcal{Set}$ spanned by the $T$-closed objects is the empty subcategory.
On the other hand, the algebras for the double powerset monad are the complete atomic boolean algebras. The category of complete atomic boolean algebras is not empty, so it can't be isomorphic to the empty subcategory of $\mathcal{Set}$.
More generally, the Eilenberg-Moore category isn't necessarily equivalent to any full subcategory of $\mathcal{C}$. For example, the category of algebras for the free group monad are the groups, but the category of groups can't be equivalent to a full subcategory of $\mathcal{Set}$.
An easy way to see that is that the automorphism groups of objects are the same in any full subcategory (due to fullness). The automorphism groups of objects of $\mathcal{Set}$ are the symmetric groups. But the category of groups contains $\mathbb{Z}_5$, whose automorphism group is $\mathbb{Z}_4$, which isn't a symmetric group.
Best Answer
Since every algebra is a coequalizer of free algebras, if $J$ is a left adjoint and the Eilenberg-Moore category is cocomplete, then $J$ is an equivalence. This follows from the fact that coreflective subcategories of cocomplete categories are closed under colimits. It is very common for the Eilenberg-Moore category to be cocomplete-for instance it suffices that $\mathcal A$ be locally presentable and $T$ preserve sufficiently filtered colimits, or that every epimorphism splits in $\mathcal A$, such as for sets.
EDIT: As Arnaud points out, cocompleteness is not necessary here-the canonical coequalizer in the EM category will be reflected back into the Kleisli category whether or not other colimits exist. So it’s necessary and sufficient that $J$ be an equivalence.