Hopefully I can expand on Kevin Arlin's comments in a helpful way.
The answer is yes, there is a Kleisli adjunction.
Preamble
I'll copy the nLab definitions to stay self-contained.
Let $K$ be a 2-category, $t:a\to a$ a monad, $(a_t,f_t,\lambda)$ a Kleisli object for $t$, meaning a representing object for the functor $K\to \newcommand\Cat{\mathbf{Cat}}\Cat\newcommand\oppd{\operatorname{.}}$ that sends an object $x$ to the right $t$-modules on $x$, $\newcommand\RMod{\operatorname{RMod}}\RMod(x,t)$.
So $a_t$ is a 0-cell, $f_t:a\to a_t$ a 1-cell, and $\lambda:f_tt\to f_t$ a 2-cell, such that
for any right module $(r,\alpha)$, with $r:a\to x$, $\alpha : rt\to r$, there is a unique morphism $a_t\to x$ whose composite with $f_t$ (resp. $\lambda)$ is $r$ (resp. $\alpha$).
Edit: An elementary reformulation of the definition of a Kleisli object:
A Kleisli object for a monad $(a,t:a\to a,\mu:t^2\to t,\eta : 1_a\to t)$ consists of the data of a 0-cell $a_t$, and a right $t$-module $(f_t : a\to a_t,\lambda : f_tt\to f_t)$ on $a_t$ such that the following universality conditions are satisfied.
Object condition: For any right $t$-module on $x$, $(r:a\to x, \alpha : rt\to r)$, there is a unique morphism $g : a_t\to x$ such that
$(r,\alpha) = (gf_t, g\oppd \lambda)$.
Morphism condition: For two right $t$-modules on $x$, which we know are of the form $(gf_t,g\oppd\lambda)$ and $(hf_t,h\oppd\lambda)$ by the object condition, for $g,h:a_t\to x$ and for every morphism of right $t$-modules $\beta: gf_t\to hf_t$, there is a unique 2-cell $\gamma : g\to h$ such that $\beta = \gamma\oppd f_t$.
The adjoint 1-cells:
We already have $f_t:a\to a_t$, so we need $g_t:a_t\to a$, which should correspond to a right $t$-module structure on $a$. Luckily, we already have a canonical one, $(t,\mu)$, where $\mu:t^2\to t$ is the multiplication of the monad. Thus we get a map $g_t$ from
the universal property, such that
$g_tf_t=t$ and $g_t\oppd\lambda = \mu$.
The unit:
Then the unit of the monad, $\eta:1_a\to t=g_tf_t$ is the unit of the adjunction.
Constructing the counit:
To construct the counit, $\epsilon : f_tg_t\to 1_{a_t}$, we need to understand
$f_tg_t : a_t\to a_t$. However, since $a_t$ represents right modules, this morphism classifies the right module on $a_t$, $(f_tg_tf_t,f_tg_t\oppd\lambda)$, but by definition of
$g_t$, this is equal to $(f_tt,f_t\oppd\mu)$.
Similarly, $1_{a_t}$ corresponds to the module $(f_t,\lambda)$.
Now you can check that $\lambda: f_tt\to f_t$ is a morphism of right $t$-modules between these two,
since
$$
\require{AMScd}
\begin{CD}
f_ttt @>f_t\oppd\mu>> f_tt \\
@V\lambda\oppd t VV @VV\lambda V\\
f_tt @>\lambda>> f_t \\
\end{CD}
$$
commutes, because this diagram is one of the diagrams that are required for $\lambda$ to be a multiplication making $f_t$ a $t$-module in the first place.
Thus $\lambda$ induces a morphism $\epsilon : f_tg_t\to 1_{a_t}$ satisfying $\epsilon\oppd f_t = \lambda$.
The triangle identities:
For the triangle identities, we now have
$$(\epsilon\oppd f_t)(f_t\oppd \eta) = \lambda(f_t.\eta)=1_{f_t}$$ by the unit axiom of $\lambda$. For the other, we can understand
$$(g_t\oppd \epsilon)(\eta \oppd g_t) : g_t\to g_t $$
by composing with $f_t$ to get the corresponding endomorphism of the right $t$-module $(t,\mu)$.
$$((g_t\oppd \epsilon)(\eta\oppd g_t))\oppd f_t
= (g_t\oppd \epsilon \oppd f_t)(\eta\oppd g_t\oppd f_t)
= (g_t\oppd \lambda)(\eta\oppd t)
= \mu(\eta\oppd t)
= 1_t,
$$
by the unit axiom of $\mu$.
Since $1_t = 1_{g_t}\oppd f_t$, we conclude
$$(g_t\oppd \epsilon)(\eta\oppd g_t) = 1_{g_t},$$
as desired.
Best Answer
So, this is not going to be a particularly enlightening example, but we can definitely walk through it together. In the interest of showing something a bit more complex as well, I've also included a worked version of another pair of adjoints.
So you're totally right. Every groupoid is a category, so there is an inclusion functor
$$\iota : \mathsf{Grpoid} \to \mathsf{Cat}$$
Moreover, by ignoring any non-isos, we can look at "the biggest groupoid living inside $\mathsf{Cat}$". This gives us a functor
$$\text{core} : \mathsf{Cat} \to \mathsf{Grpoid}$$
The first question to ask yourself after "is this an adjunction?" is "which one goes on which side?". Since a functor sends isos to isos, the image of any groupoid under a functor must be a groupoid itself. So a functor $\iota G \to C$ must factor through the core of $C$ (which is the largest possible groupoid in $C$). Succinctly:
$$\iota \dashv \text{core}$$
So now we want to look at the monad associated to this adjunction. I always remember this in terms of the free-forgetful adjunction of an algebraic structure. For groups, say, we have
$$(F : \mathsf{Set} \to \mathsf{Grp}) \dashv (U : \mathsf{Grp} \to \mathsf{Set})$$
This should give us a monad on $\mathsf{Set}$, so our monad must be $UF : \mathsf{Set} \to \mathsf{Set}$.
By analogy, our monad is
$$\text{core} \circ \iota : \mathsf{Grpoid} \to \mathsf{Grpoid}$$
And now we see why this is not a particularly enlightening example: If we look at the largest groupoid contained in $G$.... Well, $G$ is already a groupoid. So we're actually looking at the identity monad $I$!1
Of course, we can still work out the things that you've asked for.
As a slightly more instructive example, here is something that I thought about a little while ago:
Given a graph $\Gamma = (V,E)$, we can form the Right Angled Artin Group
$$A\Gamma = \langle V~|~[x,y] \iff xEy \rangle$$
This is the free group on the vertices, where we force two vertices to commute when they're connected by an edge.
Conversely, given a group $G$, we can form is Commutation Graph $CG$ whose vertices are group elements, and we connect two group elements exactly when they commute.
You can convince yourself that these form a pair of adjoint functors $A \dashv C$ between $\mathsf{ReflGph}$ and $\mathsf{Grp}$. Here $\mathsf{ReflGph}$ is the category of graphs (sets equipped with a reflexive, symmetric binary relation). We really need reflexivity here (do you see why?) but that's fine. Graph homomorphisms are the natural thing: Functions on the vertices which preserve (but don't necessary reflect) the edge relation.
Now I claim $A \dashv C$ is a pair of adjoint functors. So then the monad will be $CA : \mathsf{ReflGph} \to \mathsf{ReflGph}$. It sends a graph $\Gamma$ to a new graph $CA\Gamma$ with
a vertex for every element of $A\Gamma$. So we roughly have a vertex for each word in the vertices of $\Gamma$, but we have to declare certain words equivalent (like $xy$ and $yx$ for adjacent $x,y \in \Gamma$).
an edge connecting any two commuting elements of $A \Gamma$. So, for instance, we have a countable clique connecting all of the $x^n$ for $n \in \mathbb{Z}$. Moreover, we have an edge from the empty word to every other vertex.
Ok. So this is a monad on $\mathsf{ReflGph}$. What are the operations?
The product should send $CACA \Rightarrow CA$. So we need to know how to map the vertices of $CACA\Gamma$ to vertices of $CA\Gamma$. But vertices of $CACA\Gamma$ are words in the vertices of $CA\Gamma$. That is, words of words of vertices of $\Gamma$. So concatenation gives us a map from $CACA\Gamma \to CA\Gamma$. This may sound complicated at first, but after some meditation it really is simple: Given a word of words, say $[[x,y],[z],[w,x^{-1},y^{-1}]]$ we concatenate these to get the word $[x,y,z,w,x^{-1},y^{-1}]$.
The unit, though, is simple even on first hearing! $\Gamma$ sits naturally as a subgraph of $CA\Gamma$ by sending each vertex to the word of length $1$ containing that vertex.
1: As an aside, this situation happens often enough to have a name. When the inclusion functor $\mathcal{C} \hookrightarrow \mathcal{D}$ admits a right adjoint we call $\mathcal{C}$ a coreflective subcategory of $\mathcal{D}$.
I hope this helps ^_^