a coalgebra over a field $k$ is a vector space $C$ over $k$ together with $k$-linear maps $$\text{comultiplication } \Delta: C \to C\otimes_k C \text{ and} $$ $$\text{counit } \epsilon: C \to k$$ such that
$$(\mathrm{id}_C \otimes \Delta) \circ \Delta = (\Delta \otimes \mathrm{id}_C) \circ \Delta;$$
$$(\mathrm{id}_C \otimes \epsilon) \circ \Delta = \mathrm{id}_C = (\epsilon \otimes \mathrm{id}_C) \circ \Delta.$$
A coalgebra $C$ is called coaugmented if there exists another $k$-linear map $u: k\to C$ such that $\epsilon\circ u=\mathrm{id}_k$.
For an coaugmented coalgebra $(C,k,\Delta,\epsilon, u)$, let $\bar{C}$ denote $\ker \epsilon\subset C$. It is clear that $C\cong \text{Im}(u)\oplus \ker \epsilon$ as a $k$-linear space. Moreover we can prove that for any $x\in \bar{C}$ we have $$ \Delta(x)=x\otimes u(1_k)+u(1_k)\otimes x +\sum x^{\prime}\otimes x^{\prime\prime} $$ where $x^{\prime}\otimes x^{\prime\prime}\in \bar{C}\otimes_k \bar{C}$. See Lemma 2.4 of this paper.
With the above result, it is natural to define the reduced comultipliciation $\bar{\Delta}:\bar{C}\to \bar{C}\otimes_k\bar{C}$ as $$ \bar{\Delta}(x):=\Delta(x)-x\otimes u(1_k)-u(1_k)\otimes x. $$
Now I want to prove that $\bar{\Delta}$ is coassiciative, i.e. $$ (\mathrm{id}_\bar{C} \otimes \bar{\Delta}) \circ \bar{\Delta} = (\bar{\Delta} \otimes \mathrm{id}_\bar{C}) \circ \bar{\Delta}. $$
By routine computation, it reduced to the identity $$ \Delta(u (1_k))=u (1_k)\otimes u (1_k). $$ Using Properties 1 and 2 above I can show that $\Delta(u (1_k))$ has no component in $\text{Im}(u)\otimes_k \bar{C}$ and $\bar{C}\otimes_k \text{Im}(u)$ and its component in $\text{Im}(u)\otimes_k \text{Im}(u)$ is $u (1_k)\otimes u (1_k)$. However I don't know how to prove $\Delta(u (1_k))$ has no component in $\bar{C}\otimes_k\bar{C}$, although it is true for the easies case: the tensor coalgebra.
It will be appreciated if you can let me know how to prove the coassociativity or if my argument is wrong.
(Modified and Updated) For your last concern, $u$ should be a coalgebra map. Guess this is why your computation does not follow. In standard references like Loday's Algebraic Operad and Peter May's More Concise Algebraic Topology, the coaugmentation map is always a coalgebra homomorphism. It follows $u(1)$ is also a group-like element, since $u$ is an coalgebra map, so $\Delta(u(1)) = (u \otimes u)(\Delta_k(1)) = u(1) \otimes u(1)$ and $\epsilon_C(u(1)) = \epsilon_k(1) = 1$). $u(1) \not\in \ker\epsilon$.
Back to your main question, this is straightforward computation without any trick and this has been more or less done in Augmented coalgebras. The key in that computation is that $1$ is an group-like element. I thought the short comment that $u(1)$ is group-like should solve your question already. Looks like I need to give a detail answer in the answer section.
Here we assume $g$ is any group-like element and $\Delta$ is coassociative. Define $\bar\Delta(x) = \Delta(x) - g \otimes x - x \otimes g$. I assume the Sweddler notation.
$(\overline{\Delta} \otimes Id)\overline{\Delta}(x) = x_{(11)} \otimes x_{(12)} \otimes x_{(2)} - x_{(1)} \otimes g \otimes x_{(2)} - g \otimes x_{(1)} \otimes x_{(2)} - x_{(1)} \otimes x_{(2)} \otimes g + x \otimes g \otimes g + g \otimes x \otimes g + g \otimes g \otimes x$ $(Id \otimes \overline{\Delta})\overline{\Delta}(x) = x_{(1)} \otimes x_{(21)} \otimes x_{(22)} - x_{(1)} \otimes x_{(2)} \otimes g - x_{(1)} \otimes g \otimes x_{(2)} + x \otimes g \otimes g -g \otimes x_{(1)} \otimes x_{(2)} + g \otimes x \otimes g + g \otimes g \otimes x $
$(\overline{\Delta} \otimes Id)\overline{\Delta}(x) - (Id \otimes \overline{\Delta})\overline{\Delta}(x) = x_{(11)} \otimes x_{(12)} \otimes x_{(2)} - x_{(1)} \otimes x_{(21)} \otimes x_{(22)} = 0$ since $\Delta$ coassociative.
Update: Here Well-definedness of reduced coproduct details on well-definedness of reduced comultiplication for your last concern.