I would like to propose to you the following problem and my proposed solution. In particular, I am unsure in how to correctly characterize the Walrasian demand. Can you please have a look at it and express your opinions and corrections?
Let ${x}=(x_1,x_2)$ denote the consumption vector, ${p}=(p_1,p_2)$ denote the price vector, and let $w$ be the consumer's wealth. The utility maximization problem is
\begin{equation*} \max_{{x\geq 0}} \ \ (x_1 + 1)^\alpha (x_2+1)^\beta \ \ \ \text{ s.t. } {p \cdot x} \leq w.\end{equation*}
with $\alpha + \beta = 1$ with $\alpha, \beta > 0$. The Lagrangian function for the UMP is
\begin{equation*} \mathcal{L} ({x}, \lambda, {\mu} ) = (x_1 + 1)^\alpha (x_2+1)^\beta- \lambda ({p \cdot x} -w) + {\mu \cdot x},\end{equation*}
where $\lambda$ and ${\mu} = (\mu_1,\mu_2)$ are the Lagrangian multipliers.
The Kuhn-Tucker necessary conditions for the consumer problem are formed by the following conditions:
\begin{equation*} \begin{aligned} \frac{\partial \mathcal{L}}{\partial x_1} = \alpha(x_1+1)^{\alpha-1} (x_2 +1)^\beta - \lambda p_1 + \mu_1&= 0,\\[5pt] \frac{\partial \mathcal{L}}{\partial x_2} = \beta(x_2+1)^{\beta-1} (x_1 +1)^\alpha - \lambda p_2 + \mu_2 &= 0, \end{aligned} \end{equation*} with $\lambda \geq 0$ and ${\mu} \geq 0$; the initial constraints,
\begin{equation} {p \cdot x} \leq w \ \ \ \text{ and } \ \ \ {x } \gg 0,\end{equation}
and the complementary slackness conditions,
\begin{equation*} \lambda \frac{\partial \mathcal{L}}{\partial \lambda} = \lambda ({p \cdot x} - w)=0 \ \ \ \ \text{ and } \ \ \ \ \mu_i \frac{\partial \mathcal{L}}{\partial \mu_i} = \mu_i x_i = 0 \ \text{ for } i=1,2. \end{equation*}
Note that under these conditions $\partial u({x})/\partial x_i \leq \lambda p_i$. Since $\nabla u({x}) \gg 0$ and ${p} \gg 0$, this implies $\lambda \geq [\partial u({x})/\partial x_i]/p_i > 0$. In words, the Lagrangian multiplier $\lambda$ is positive and the constraint is binding. Therefore, by $\lambda ({p \cdot x} - w)=0$, we can conclude that ${p \cdot x} = w$ (Walras' law holds).
Interior solutions are of the type ${x} \gg 0$ which imply ${\mu} = {0}$. As we have shown above, because $u({x})$ is an increasing function and since we assume ${p} \gg 0$, Walras' law holds. Thus, the Kuhn-Tucker conditions restrict to the system of equations
$$\begin{align} &\alpha(x_1+1)^{\alpha-1} (x_2 +1)^\beta = \lambda p_1, \qquad &(1)\\[5pt] &\beta(x_2+1)^{\beta-1} (x_1 +1)^\alpha = \lambda p_2, \qquad &(2)\\[5pt] &{x} \gg 0, \qquad &(3)\\[5pt] &{p \cdot x} = w. \qquad &(4) \end{align}$$
By dividing (1) by (2) we obtain the key (tangency) condition for an optimum: the marginal rate of substitution of good 1 with good 2 at the optimum must equal the price ratio of the two goods
\begin{equation*} \frac{\alpha(x_1+1)^{\alpha-1} (x_2 +1)^\beta}{\beta(x_2+1)^{\beta-1} (x_1 +1)^\alpha}= \frac{p_1}{p_2} \end{equation*}
\begin{equation} \frac{\alpha (x_2 +1)}{\beta(x_1+1)}= \frac{p_1}{p_2} .\qquad (5)\end{equation}
Solving (5) for $x_2$ allows to rewrite the necessary (and sufficient) condition for ${x}=(x_1,x_2) $ to be an optimum in a useful way for our computations
\begin{equation} x_2= \frac{\beta p_1}{\alpha p_2}(x_1+1) -1. \qquad (6) \end{equation}
By substituting (6) into (4) we are now able to solve for $x_1 ({p}, w)$
$$\begin{align} &p_1 x_1 +p_2 \bigg(\frac{\beta p_1}{\alpha p_2}(x_1+1) -1\bigg) = w \nonumber \\[5pt] &p_1 x_1 = \bigg(w - \frac{\beta }{\alpha}p_1 + p_2 \bigg)\frac{\alpha}{\alpha + \beta} \nonumber \\[5pt] &x_1 ({p}, w) = \frac{\alpha (w+p_2)}{p_1} - \beta , \qquad \qquad (7) \end{align}$$
and by substituting (7) into (6) we cans solve for $x_2 ({p}, w)$ \begin{equation*} x_2 ({p}, w) = \frac{\beta (w+p_1)}{p_2} - \alpha .\end{equation*}
For boundary solutions we need to look at the cases where $x_1 = 0$ or $x_2 = 0$ (having both equal zero is uninteresting and certainly not the case for any locally nonsatiated function). The complementary slackness condition $\mu_i x_i = 0 $ for $i=1,2$ implies in the first case $\mu_1 \geq 0$ and in the second $\mu_2 \geq 0$. As before, because $u({x})$ is an increasing function and since we assume ${p} \gg 0$, Walras' law holds (in both cases). \[-7pt]
In the first case, $x_1 = 0$ implies $\mu_1 \geq 0$. Thus, the Kuhn-Tucker conditions restrict to the system of equations
$$\begin{align} &\alpha(x_1+1)^{\alpha-1} (x_2 +1)^\beta \leq \lambda p_1, \qquad &(8)\\[5pt] &\beta(x_2+1)^{\beta-1} (x_1 +1)^\alpha = \lambda p_2,\qquad &(9)\\[5pt] &x_1 = 0, \ \ x_2 > 0, \qquad &(10)\\[5pt] &{p \cdot x} = w. \qquad &(11) \end{align}$$
By substituting the equation in (10) into the budget constraint (11) we obtain the candidate for an optimum, $x_2=w/p_2$. Then, if we divide (8) by (9), and consider $x_1 = 0$, we obtain the necessary (and sufficient) condition for $x_2=w/p_2$ being optimal\footnote{Note that, for boundary solutions, the indifference curve need not be tangent to the budget line.}
\begin{equation*} \frac{\alpha (x_2 +1)}{\beta(x_1+1)} \leq \frac{p_1}{p_2} \end{equation*}
\begin{equation} \frac{\alpha (x_2 +1)}{\beta} \leq \frac{p_1}{p_2}. \end{equation}
Since $\alpha (x_2 +1) / \beta > 0$, there exists some price vector ${p \gg 0}$ that satisfies (12). Hence $x_2=w/p_2$ is optimal for those price vectors ${p}$ that satisfy (12).
In the second case, $x_2 = 0$ implies $\mu_2 \geq 0$. This time we omit most of the algebra and explanation because they bear close similarity to the case above. Jumping to the conclusion: by the Walras' law the candidate for the optimum is $x_1=w/p_1$ and the necessary condition for it being an optimum is
\begin{equation} \frac{\alpha}{\beta(x_1 +1)} \geq \frac{p_1}{p_2}. \end{equation}
Since $\alpha / (x_1 +1)\beta > 0$, there exists some price vector ${p \gg 0}$ that satisfies (12). Hence $x_1=w/p_1$ is optimal for for price vectors ${p}$ that satisfy (13).
Finally, the Walrasian demand in compact notation is
\begin{equation} x({p}, w) = \begin{cases} \bigg(\dfrac{w}{p_1}, 0\bigg) \quad &\text{ if } \ \dfrac{p_1}{p_2} \leq \dfrac{\alpha}{\beta(x_1 +1)} \\[5pt] \bigg(\dfrac{\alpha (w+p_2)}{p_1} - \beta, \, \dfrac{\beta (w+p_1)}{p_2} - \alpha \bigg) \quad &\text{ if } \ \dfrac{\alpha}{\beta(x_1 +1)} < \dfrac{p_1}{p_2} < \dfrac{\alpha (x_2 +1)}{\beta}\\[5pt] \bigg(0,\dfrac{w}{p_2}\bigg) \quad &\text{ if } \ \dfrac{p_1}{p_2} \geq \dfrac{\alpha (x_2 +1)}{\beta} \end{cases} \end{equation}
