When we trace out system b, what we are doing is basically reducing the system down to as if we had just measured system a
Its as if you had just measured or discarded system $b$.
Otherwise yes, the probability distribution over the computational basis states described by $\rho_a$ on $\mathcal{H}_a$ is precisely the marginal of the distribution described by $\rho_{ab}$ on $\mathcal{H}_a \otimes \mathcal{H}_b$. For example, if each system consists of a single qubit, we can write
$$
\rho_{ab} = \sum_{i,j,k,\ell=0}^1 \rho_{ij,k\ell}|ij\rangle \langle k\ell|
$$
If you were to measure this in the computational basis, you would end up with a joint probability distribution of both bit outcomes over $\{00, 01, 10, 11\}$ given by $\text{diag}(\rho) = (\rho_{00,00}, \rho_{01,01}, \rho_{10, 10}, \rho_{11, 11})$. If we treat the measurement of the first system as a random variable $A\in\{0,1\}$ with an associated distribution $p_A$, we can compute the marginal probability for measuring bit "$i$" in system $a$ as:
\begin{align}
p_A(i) &= \sum_{b\in\{0,1\}} p(A=i, B=b) \\
&= \sum_{b\in\{0,1\}} \rho_{ib, ib}
\end{align}
And so the full distribution $p_A$ is given by
\begin{align}
p_A(0) &= \rho_{00,00} + \rho_{01,01} \\
p_A(1) &= \rho_{10,10} + \rho_{11,11} \\
\end{align}
Compare this to a known formula for $\text{Tr}_B (\rho_{ab})$ (for example, this question):
$$
\text{Tr}_B(\rho_{ab}) = \begin{pmatrix}
\rho_{00,00} + \rho_{01,01} & \rho_{00,01} + \rho_{01,11} \\
\rho_{10,00} + \rho_{11,01} & \rho_{10,10} + \rho_{11,11}
\end{pmatrix}
$$
The diagonal elements match the marginal probabilities computed above. If you lift the restriction that systems $a$ and $b$ are qubits and allow them to be $d$-level systems you can repeat the same calculations to show that the computational measurement probability distribution for $\text{Tr}_B (\rho_{ab})$ is the marginal probability distribution of $\rho_{ab}$ for a bipartite system with arbitrary dimensions.