Lax cones for a monad A monad $T$ on $\cal C$ is a lax functor $\mathbb T : \mathbf{1}\to\bf Cat$. The lax colimit and limit of $\mathbb T$ are the Klesili and EM categories ${\cal C}^{\mathbb T}$, ${\cal C}_{\mathbb T}$ of the monad.
Hence, it seems reasonable for the category of lax co/cones for $\mathbb T$   to be equivalent to the category whose objects are adjunctions $F\dashv G$ splitting the given monad as a composition $GF$.
Writing down the definition of a lax cone for $\mathbb T$, one obtains a functor $S\colon {\cal X}\to \cal C$ with a natural transformation $\sigma \colon TS\to S$ satisfying certain commutativities:
$$
\begin{array}{lr}
\begin{array}{ccc}
TTS &\to & TS \\
\downarrow && \downarrow \\
TS &\to& S
\end{array}
&
\begin{array}{ccc}
S &\to &TS\\
&\searrow&\downarrow\\
&& S
\end{array}\\
&&\\
\sigma\circ \mu S = \sigma\circ T\sigma, & \sigma \circ \eta S = 1_S
\end{array}
$$
This means that morally $\sigma = S\epsilon$, for $\epsilon\colon LS\to 1$ the counit of the wannabe adjunction (the equality $\sigma \circ \eta S = 1_S$ is hence half of the zigzag identities). 
It seems now unlikely that the functor $S\colon \cal X\to C$, the right adjoint in the wannabe adjunction $L\dashv S$, becomes automatically continuous. So,

is the conjecture above false? What should it be replaced by?

 A: I believe that your claim is false. 
Consider $\mathcal X=1$ then a pair $\langle S,\sigma \rangle$ as you describe (a cone for $\mathbb T$) should be nothing but:


*

*an object $S(*)$, where $*$ is the only object in $1$;

*a $T$-algebra structure over $S(*)$, namely $\sigma_* \colon TS(*) \to S(*)$.


The left adjoint for $S$ should be an $L \colon \mathcal C \to 1$ and there are not many functors of this kind, just the constant one. 
Similarly $\epsilon \colon LS \to 1_{1}$, the counit, is given by the only morphism $\epsilon_* \colon LS(*)=* \to *$ which must be $1_{*}$, the only morphism in $1$.
If your claim was true, that is $S\epsilon=\sigma$, this would imply that $$\sigma_*=S(\epsilon_*)=S(1_*)=1_{S(*)}\ .$$
This is basically saying that every $T$-algebra should have the identity as underlying structure map. This is not clearly the case because:


*

*for start there is no reason why $TS(*)=S(*)$ ($TS(*)$ is the source of $\sigma$, while $S(*)$ is the source of $1_{S(*)}$);

*there are lots of monads whose algebras have structure maps that are not the identity.

