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=:b_1 \\
\implies a'y&\le b'-a'x_0=:b'_1
\end{align}
where $b_1>0$ and scalar $b'_1>0$. Writing the matrices in the entry form so as to be clear, we have
\begin{align}
\sum_j\frac{A_{ij}}{b_{1,i}}y_j&\le1 \\
\implies \sum_j\frac{a'_j}{b'_1}y_j&\le1.
\end{align}
By Farkas' lemma or the separating hyperplane theorem, $\sum_j\frac{a'_j}{b'_{1,i}}$ is a convex combination of $\frac{A_{ij}}{b_{1,i}}$, or $\exists u\in R^n$ such that $u_i\ge0\, \forall i, \sum_iu_i=1$ and
$$\frac{a'_j}{b'_1}=\sum_i u_i\frac{A_{ij}}{b_{1,i}}$$
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 \lambda_i b_i  \tag2
\end{align}
for some $\lambda_i\ge0, \forall i$ (without requiring $\sum_i\lambda_i=1$).

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I doubt you can derive (2) from (1) directly without other conditions. It is pretty easy to find a counterexample even for $x_0=0$.