This is not a proof per se but for detailed discussion with Dima Pasechnik. It is too long for the comment section.

Suppose $P$ is bounded and not a singleton. The original proposition does not stipulate $0\in $ interior of $P$.

I transcribe Dima Pasechnik's answer in detail as follows. We pick an interior point $x_0$ of $P$, i.e., $Ax_0<b$. Let $y:=x-x_0$.
\begin{align}
Ay&\le b-Ax_0=:\beta \\
\implies a'y&\le b'-a'x_0=:\beta'
\end{align}
where vector $\beta>0$ and scalar $\beta'>0$. Writing the matrices in the entry form so as to be clear, we have
\begin{align}
\sum_j\frac{A_{ij}}{\beta_i}y_j&\le1 \\
\implies \sum_j\frac{a'_j}{\beta'}y_j&\le1.
\end{align}
By Farkas' lemma or the separating hyperplane theorem, $\forall j$, $\frac{a'_j}{\beta'}$ is a convex combination of $\Big\{\frac{A_{ij}}{\beta_i}\Big\}_i$, or $\exists u\in R^n$ such that $u_i\ge0\, \forall i, \sum_iu_i=1$ and
$$\frac{a'_j}{\beta'}=\sum_i u_i\frac{A_{ij}}{\beta_i}$$
or
$$\frac{a'_j}{b'-\sum_ja'_jx_{0,j}}=\sum_i u_i\frac{A_{ij}}{b_i-\sum_jA_{ij}x_{0,j}} \tag1$$
$\forall j$.

Now, the question is how one derives from Equation (1) the desired inequalities in the question, i.e.
\begin{align}
a'_j &= \sum_i\lambda_i A_{ij} \quad\forall j, \\
b' &\geq \sum_i\lambda_i b_i \tag2
\end{align}
for some $\lambda_i\ge0, \forall i$ (without requiring $\sum_i\lambda_i=1$).

Note that the convex combination in (1) is for the ratio as opposed to for the numerator and denominator separately. Moreover, the denominator of (1) involves $A, a', x_0$ rather than just $b$ and $b'$, while (2) especial the second relation $b'\ge \lambda^Tb$ is independent of $A, a'$ especially $x_0$.

I doubt you can derive (2) from (1) directly without other conditions.

If we define as Dima Pasechnik suggests
$$\lambda_i := u_i\frac{\beta'}{\beta_i},$$
we obtain the equation above Inequality (2) which I had obtained without the normalization or the dual polytope concept, as well as for $x_0=0$ which makes $\beta=b$ and $\beta'=b'$ the equality version of (2). However, it is not clear how we could arrive at Inequality (2) under the general condition where both $\beta$ and $\beta'$ depend on $x_0$ which is the ultimate question I am after, since $\beta$ and $\beta'$ depend on $x_0$ while $b$ and $b'$ do not.