Upper densities
In the following, I freely use the well known fact that $s_\lambda$-Hausdorff measure gives mass $2^{-k}$ to all intervals that make up the stage $k$ in the construction of $C_\lambda$.
I don't think it is correct that $\Theta^{* s_\lambda}(C_\lambda,x)\ge c$ for all $x\in C_\lambda$ and some $c>2^{-s_\lambda}$.
For example, let $\lambda=1/3$ so that $C=C_\lambda$ is the middle-thirds Cantor set. Write $s=s_{1/3}=\tfrac{\log 2}{\log 3}$, and $$ D(x,r) = \frac{\mathcal{H}^{s}(B(x,r)\cap C)}{(2r)^s} $$
By self-similarity, $D(0,r)=D(0,r/3)$, so $$ \Theta^{*s}(C,0) = \max\{ D(0,r): r\in [1/3,1]\}. $$ Now if $r\in [1/3,2/3]$, then $\mathcal{H}^s(B(0,r))=1/2=(1/3)^s$ so $D(0,r)\le 2^{-s}$. Otherwise, write $r=2/3+t$ for some $t\in [0,1/3]$. Then $$ \mathcal{H}^s(B(0,r)\cap C) = \mathcal{H}^s([0,1/3]\cap C)+\mathcal{H}^s([2/3,2/3+t]\cap C)\le \tfrac{1}{2}+t^{-s}, $$ whence $$ D(0,r) \le \frac{\tfrac{1}{2}+t^{s}}{2^s (2/3+t)^s} \le 2^{-s}, $$ from elementary calculus. Hence $\Theta^{*s}(C,0)=2^{-s}$. There's nothing special about $\lambda=1/3$ here. Also, although the point $0$ is special, the same holds for any ternary point.
Lower densities
The lower density problem was essentially solved, for more general self-similar sets, in [Feng, De-Jun. Exact packing measure of linear Cantor sets. Math. Nachr. 248/249 (2003), 102--109]. In Theorem 1.1, a formula is given for the infimum of $$ \frac{\mathcal{H}^{s_\lambda}(B(x,r)\cap C_\lambda)}{(2r)^s} $$ over all $x\in C_\lambda$ such that $B(x,r)\subset [0,1]$. Note that the example after Theorem 1.1 are exactly the central Cantor sets (with $\lambda=(1-\beta)/2$). It seems this value is strictly larger than $2^{-s_\lambda-1}$ for all $\lambda\in (0,1/2)$.
In Theorem 2.1, Feng shows that $\Theta_*^{s_\lambda}(C_\lambda,x)$ equals this infimum for almost all $x$. Clearly, this infimum equals the smallest possible value of $\Theta_*^{s_\lambda}(C_\lambda,x)$, at least if we exclude $x=0,1$ (and if $\lambda\le 1/3$ this restriction is not necessary, as any extreme point of a construction interval will have the same density as $0$ and $1$).
Edit (after Paul-Benjamin's comment): to give an upper bound for the lower density at every point of $C_\lambda$, I consider again the case $\lambda=1/3$ for concreteness, although a similar argument should work for any $\lambda$. Suppose first that $x$ is such that it belongs to infinitely many "left intervals" and infinitely many "right intervals" in the construction (or in other words the ternary expansion of $x$ has infinitely many zeros and twos). Then there are infinitely many $k$ such that the distance from $x$ to the boundary of the level $k$ interval of the construction of $C$ containing $x$ is at least $(2/9) 3^{-k}$. Let $r=r_k=3^{-k}(1+2/9)$. Then the ball $B(x,r)$ intersects only one interval of level $k$ in the construction, so that $$ \frac{\mathcal{H}^s(B(x,r)\cap C)}{(2r)^s} = \frac{1}{2^s(11/9)^s} < 2^{-s}. $$ Otherwise, without loss of generality, from some generation on $x$ is always on the left interval of the construction. In that case, $B(x, 2 3^{-k})$ meets only one interval of the construction for all sufficiently large $k$ and we get an even smaller upper bound for the lower density. Hence $b=2^{-s} (11/9)^{-s}$ works (I'm not claiming this is optimal, though it might be).