As Zaq mentioned in a comment, the answer to both questions are yes. I will try to outline an explanation of the first. The second follows similarly.
Let $P_{y}(x) = P(x,y) = \frac{2}{\omega_{n+1}} \frac{y}{(|x|^{2} + y^{2})^{\frac{n+1}{2}}}$.
Note that $P_{y} \in C^{\infty}(\mathbb{R})$ for all $y \neq 0$. Also, for all $y \neq 0$ we have $P_{y}(x) = y^{-n} P_{1}(y^{-1}x)$. Consequently, it can be shown that $\int_{x \in \mathbb{R}^{n}} P_{y}(x) d x = 1$ by computing that $\int_{x \in \mathbb{R}^{n}}P_{1}(x) d x = 1$. A quick aside here, since you have seen the Fourier analysis already, then we have that $\widehat{P}_{y}(0) = \int_{\mathbb{R}^{n}} P_{y}(x) d x$. So, plugging into your formula, the claim is immediate.
The three proceeding properties I mentioned about $P_{(\cdot)}$ tells us that the family of functions $\{P_{y}\}_{y > 0}$ shares many of the same properties as a smooth approximation to the identity sometimes also called a mollifier. See Chapter 4 section 1 of Evans and Gariepy's book "Measure Theory and Fine Properties of Functions" if you are unfamiliar. This text includes a very detailed proof of the commonly used properties of mollifiers.
A key property that mollifiers have that $P_{y}$ does not, is that mollifiers typically have compact support. This is not necessary, but it is quite convenient. However, we can get by with instead observing that $P_{y} \in L^{1}(\mathbb{R}^{n}) \cap L^{\infty}(\mathbb{R}^{n})$ (since $y \neq 0$, we're $C^{\infty}$ and decay like $|x|^{n+1}$). Similarly $\Delta_{x}P_{y}, \partial_{y}^{2}P_{y} \in L^{1}(\mathbb{R}^{n}) \cap L^{\infty}(\mathbb{R}^{n})$.
Now, suppose $f \in L^{\infty}(\mathbb{R}^{n})$. We note that
$$
u(x,y) = \int_{\mathbb{R}^{n}} P(x-z,y) f(z) d z = P_{y}*f(z) \quad \forall y > 0, x \in \mathbb{R}^{n}
$$
By Young's inequality, and $P_{y}, \partial_{y}^{2} P_{y}, \Delta_{x} P_{y} \in L^{1}$ it follows that the convolutions: $g*P_{t}, ~ g* \Delta_{x}P_{y}$, and $ g* \partial_{y}^{2} P_{y}$ are each absolutely convergent for all $x$ and $t$.
Therefore, if $v = (v_{1}, \dots, v_{n+1})$ is a unit vector in $\mathbb{R}^{n+1}$ with $v_{n+1} = 0$, we can compute the derivative of $u(x,y)$ in the direction of $v$ as
$$
\lim_{h \downarrow 0} \frac{u ((x,y) + hv) - u(x,y)}{h} = \lim_{h \downarrow 0} \frac{P_{y} *g(x +hv) - P_{y}*g(x+hv)}{h} = \lim_{h \downarrow 0} \left(\frac{ P_{y}(\cdot + hv) - P_{y}(\cdot)}{h} \right)*g(x) = D_{v}P_{y}*g(x).
$$
In particular, $\Delta_{x} u(x,y) = \Delta_{x}P_{y}*g(x)$. It can similarly be shown that $\partial_{y}^{2}u(x,y) = \partial_{y}^{2}P_{y}*g(x)$, so that $\Delta u(x,y) = \left( \Delta P_{y} \right)* g(x) = 0 * g(x) = 0$.
Hence, under the assumption $y > 0$, we have that $u$ is harmonic at $(x,y)$. Note, this sketch of a proof does not actually require $g \in L^{\infty}$. We used that if $g \in L^{p}$ then $P_{y} \in L^{q}$ for $p^{-1} + q^{-1} = 1$. But, $P_{y} \in L^{1} \cap L^{\infty}$ implies $P_{y} \in L^{q}$ for all $1 \le q \le \infty$, so $g \in L^{p}(\mathbb{R}^{n})$ for any $p$ suffices.
The remaining question is a local question, so you may as well assume that $g$ is continuous everywhere (and bounded!) instead of just continuous at the point $x$. The type of argument follows similarly. If you desire more details, I suggest looking at Folland's "Introduction to Partial Differential Equations" Chapter 2 Section G.
