In the question it says $``\hat U_Y = $ $$\text{Spec} \varprojlim A/I^n = \varinjlim \text{Spec}\, A/I^n"$$ but this is not correct; these two ringed spaces are not equal, and even the underlying topological spaces are not equal. For example, take $A = k[x]$ and $I = (x)$. On the left hand side $$\text{Spec} \varprojlim\, k[x]/(x)^n = \text{Spec}\, k[[x]]$$ which is Spec of a DVR and has two points $(0), (x)$. On the right hand side, $\varinjlim\, \text{Spec}\, A/I^n$ is a colimit where each topological space has just a single point $(x) \in \text{Spec}\, k[x]/x^n$. In particular, the resulting ringed space is not a scheme. (Formal schemes usually aren't schemes, just ringed spaces.)
The same goes for the example in the comments, $\text{Spec}\, \mathbb Z_p$ is Spec of a DVR, has two points, while the formal spectrum of the point $\text{Spec}\, \mathbb F_p \hookrightarrow \text{Spec}\,\mathbb Z$ is a formal scheme with one point, and global sections equal to $\mathbb Z_p$ (thus not a scheme).
Besides not being equal to Spec of their global sections at the point-set level, formal schemes also behave differently algebraically than schemes when passing to an open set. For an affine scheme with global section $f \in A$, the open set $D(f)$ has global sections $A_f$. This will not be the case for formal schemes, as the localization happens before completion. When $A = k[x,y], I = (x), f = y$ for instance, on the open set $D(y)$ you can have elements like $\sum (x/y)^n$ where the power $y$ in the denominator is unbounded. There is no such element in $k[y][[x]]_y$.
As for your first question, the formal neighborhood of a point will have a complete stalks, but when the dimension of the subscheme is higher the formal neighborhood will not be complete. Intuitively this is because the formal neighborhood construction only takes the completion in directions orthogonal to the subscheme you complete.
Going back to the $k[x,y], I=(x)$ example, the stalk is complete "with respect to $x$" but not with respect to $y$. In other words, the stalk would be $k[x,y]_{(x,y)}[[t]]/(t-x)$ which is not a complete local ring, the completion being $k[x,y]_{(x,y)}[[t,s]]/(t-x,y-s)$ .
The forgetful functor from locally ringed spaces to topological spaces is actually a left adjoint, not a right adjoint; in particular, it preserves colimits. You can easily see it is not a right adjoint because it does not preserve limits: the terminal locally ringed space is $\operatorname{Spec}(\mathbb{Z})$, but its underlying space is not the terminal topological space (a singleton).
The right adjoint to the forgetful functor is fairly complicated. Roughly speaking, it takes a topological space $X$, equips it with the constant sheaf $\mathbb{Z}$ (which gives the right adjoint if you were just taking ringed spaces, not locally ringed spaces), and then "expands out" every point of $X$ into new points for each prime ideal in the stalk in order to get a locally ringed space. For more discussion and references, see this nice answer.
For the specific example you are interested in, you can just easily directly construct the colimit. Namely, let $X$ be a singleton equipped with the sheaf of rings whose global sections are $k[[x]]$. This is a locally ringed space, and it has compatible maps from your diagram. To see that it is the colimit, let $Z$ be any locally ringed space with compatible maps from your diagram. All these maps must have image that is the same single point $z\in Z$, and we get a compatible system of maps $O_{Z,z}\to k[x]/x^n$ for each $n$. These induce a map $O_{Z,z}\to k[[x]]$, and this gives a morphism of locally ringed spaces $X\to Z$ that sends the point of $X$ to $z$ and sends a section of $O_Z$ in a neighborhood of $z$ to its image in $k[[x]]$. Moreover, it is easy to see this is the unique such morphism compatible with the maps from the diagram.
(More generally, you can explicitly construct colimits of locally ringed spaces by taking the colimit of the underlying topological space and then equipping it with essentially the only obvious sheaf of rings you can define. Of course, the details are much more complicated to check in general than in your example where everything is just a singleton.)
Best Answer
I will give a counterexample in a slightly simpler situation (two copies of $C$ instead of three) but the same argument can be made for your case.
Write $\operatorname{Sch}$ (resp. $\operatorname{Set}$) for the category of schemes (resp. sets). Pick $C=\operatorname{Spec} \mathbb C[t]$, call $p$ the origin in $C$ and put $C_p=\operatorname{Spec} \mathbb C[[t]]$. Suppose that there is a colimit $X$ in the category of schemes for the diagram $D$ formed by the two maps $C_p \times C_p \to C_p \times C$ and $C_p \times C_p \to C \times C_p$. We will derive a contradiction. There is a natural map $X \to C\times C$ by definition of the colimit. I claim that the functor of points of $X$ is the (Zariski) sheafification of the presheaf taking a scheme $T$ to the subset of $(C\times C)(T)$ consisting of morphisms $(f,g)$ such that either $f$ or $g$ factors through $C_p$. To prove this claim, call $X'$ this sheafification. Then $X'$ is the colimit of our diagram $D$ in the category of Zariski sheaves $\operatorname{Sch}^{op} \to \operatorname{Set}$. In particular, there is a natural map $X' \to X$ in this category (since $X$ is the colimit in a smaller category). Conversely, a morphism of schemes $T \to X$ gives a morphism $T \to C\times C$, which factors through $X'$ since $C_p \times C$ and $C \times C_p$ jointly surject onto $X$. So, we have $X=X'$, which proves the claim.
Now, pick an affine neighbourhood $U=\operatorname{Spec} A$ of $(p,p)$ in $X$. We can choose an affine neighbourhood $V=\operatorname{spec} B$ of $p$ in $C$ which is small enough so that $C_p \times V \to X$ and $V \times C_p \to X$ both factor through $U$. Identifying $C\times C$ with $\mathbb A^2_k=\operatorname{Spec}\mathbb C[t_1,t_2]$, we find that $A$ is a sub-$\mathbb C[t_1,t_2]$-algebra of $\mathbb C[[t_1,t_2]]$ which factors through both $\mathbb C[[t_1]]\otimes_\mathbb C B$ and $B \otimes_\mathbb C \mathbb C[[t_2]]$. Since $B$ is obtained from $\mathbb C[t]$ by inverting finitely many polynomials, $U=\operatorname{Spec} A$ is an open subscheme of $C \times C=\mathbb A^2_k$. This is incompatible with $U$ being contained in $X$, so we have our contradiction (for example because such an $U$ will contain many closed points which do not lie on one of the coordinate axes).