I do not know if this is what you are after. The following shows that the Jordan form can be replaced by the Schur form. This feels a bit like cheating. Is this enough? The trickier parts are the statements about the real parts of the
eigenvalues. It is of course a bit harder to say useful things about
eigenvalues when we are not allowed to speak about eigenvectors. If computational
considerations are a concern then this might be admissible as the Schur
form is obtained via orthogonal/unitary transformations. On the other hand in computing the Schur form you compute at least one eigenvector.

It is bit more convenient to argue over $\mathbb{C}$. Real version of what is below are easy to come by.

**Note 1:** I tried my best to avoid eigenvectors. To me this feels
artificial. The best proof I know for (3) $\Rightarrow$ (1) is:
``Assume $\lambda$ is an eigenvalue of $A$ with nonnegative real
part. Let $x$ be an eigenvector. Then $e^{At}x = e^{\lambda t}x$ does not
converge to zero, contradicting exponential stability.'' Compared with
this the proof below is a bit over the top.

**Note 2:** If computational approaches to stability are what you are
after and you do not like eigenvectors, why do you not simply dispense
with (1) and live with the equivalence of (2) and (3). The solution of (2)
asks for solving a system of linear equations. This can frequently be done
and is often much easier than computing the eigenvalues anyway.

**Finally, the answer:** Most of this is only a sketch. I hope this will be sufficient. For the equivalence (2) $\Leftrightarrow$ (3) no information
about the eigenstructure of $A$ is needed.

We call the equilibrium position $\hat x:=0$ of $\dot x = Ax$ exponentially stable, if there
are $M,\beta>0$ such that for all $t\geq0$ we have
$$ \|e^{At}\| \leq M e^{-\beta t}.$$
By equivalence of norms on finite-dimensional vector spaces, we may use any norm on $\mathbb{C}^{n \times n}$ for this.
I will allow myself the (common) abuse of terminology of calling the
system exponentially stable, if the equilibrium position $\hat x=0$ is.

(2) $\Rightarrow$ (3): Let $P$ be the solution corresponding to
$Q=I$. Consider the Lyapunov function $V(x) = x^T P x$ and compute
that the derivative along solutions is
$$ \dot V (x(t)) = - \|x(t)\|^2. $$
From this conclude exponential stability.

(3) $\Rightarrow$ (2): Given $Q$ positive definite, define (as you say)
$$ P := \int_0^\infty e^{A^*t}Q e^{At} dt .$$
Check that this is well-defined by exponential stability, Hermitian and positive definite and that the
integral evaluates to $-Q$.

(1) $\Rightarrow$ (3): We will use the following small

**Lemma:** If $M\in \mathbb{C}^{n \times n}$ defines an exponentially
stable system and $\lambda \in \mathbb{C}$ with $\mathrm{Re} \lambda <0$
then for all $K \in \mathbb{C}^{n \times
1}$ the matrix
$$
\begin{bmatrix}
M & K \\
0 & \lambda
\end{bmatrix}
$$
defines an exponentially stable system.

**Proof:** Follows by a computation using the variation-of-constants
formula. $\square$

With this lemma at hand, the argument is now as follows. Assume all
eigenvalues of $A$ have negative real part. Let a Schur form of $A$ be
given by
$$ UAU^* =
\begin{bmatrix}
\lambda_1 & * & \ldots& * \\
0 & \lambda_2 & *&*\\
& \ddots& \ddots & * \\
0 &&0& \lambda_n
\end{bmatrix}$$
The $1\times 1$ submatrix $\lambda_1$ clearly defines an exponentially
stable system. The exponential stability of the upper left $k \times k$
submatrices then follows from the lemma. By continuing until $n$ we obtain
exponential stability of $UAU^*$ and then also of $A$.

(3) $\Rightarrow$ (1): Again let $UAU^*$ be a Schur form of $A$. The
diagonal entries of $U e^{At}U^*$ are of the form $e^{\lambda_j t}$, where
$\lambda_j, j=1,\ldots,n$ are the eigenvalues of $A$. As $e^{At}$ goes to
zero exponentially as $t\to\infty$, it follows that $e^{\lambda_j t}\to 0$
exponentially for all $j$. Thus all eigenvalues have negative real part.