If I include $z_1$ in the model, like this:
$$
> y = \beta_0 + \beta_1 x + \beta_2 z_1 + e,
> $$
Does that mean that $\beta_1$ is predominantly capturing the effect of $z_2$?
Yes. This can be seen using the Frish-Waugh-Lovell theorem:
If you regress:
$$
y = \beta_0 + \beta_1 x + \beta_2 z_1 + e,
$$
then $\beta_1$ will be the same as the corresponding coefficient of a modified regression:
$$
\hat y = \gamma_0 + \beta_1 \hat x + \hat e \tag{1}
$$
where $\hat x$ is the residual from regressing $x$ on $z_1$ and the same for $\hat y$. Now if we regress $x$ on $z_1$ then the residual is equal to:
$$
M_{z_1} x,
$$
where $M_{z_1} = 1 - z_1(z_1'z_1)^{-1}z_1'$ is the annihilator matrix. If $x = z_1 + z_2$ then:
$$
M_{z_1} x = (1 - z_1(z_1'z_1)z_1')(z_1 + z_2) = M_{z_1}z_2
$$
As such, substituting in $(1$), we have:
$$
\hat y = \gamma_0 + \beta_1 \hat z_2 + \hat e,
$$
where $\hat z_2$ is now the residual from regressing $z_2$ on $z_1$. Using the Frish-Waugh-Lovell theorem, in the reverse direction, this gives that $\beta_1$ is also equal to the coefficient in the following regression:
$$
y = \delta_0 + \beta_1 z_2 + \delta_2 z_1 + \varepsilon.
$$
In other words, $\beta_1$ will also be equal to the coefficient for $z_2$ for a regression of $y$ on both $z_2$ and $z_1$. Notice, however, that in general $\beta_2 \ne \delta_2$ (so the coefficients for $z_1$ will not be equal in the two regressions).
Another way to see this is by immediately substituting $x = z_1 + z_2$ into the regression $(1)$ then:
$$
\begin{align*}
y &= \beta_0 + \beta_1(z_1 + z_2) + \beta_2 z_1 + e,\\
&= \beta_0 + \beta_1 z_2 + (\beta_1 + \beta_2) z_1 + e.
\end{align*}
$$
So the coefficient on $z_2$ in the new regression is identical to the coefficient on $x$ in the original one $(\beta_1)$ , while the coefficient on $z_1$ in the new regression is the sum of the coefficients on $x$ and $z_1$ in the original regression $(\beta_1 + \beta_2)$.
And a follow up question is: if $z_1$ is correlated with the error term, will including it in the model bias the estimate of $\beta_1$?
The opposite is true. Including $z_1$ into the regression will make the estimate of $\beta_1$ unbiased. Consider the following data generating process:
$$
y = \beta_0 + \beta_1 x + e, \tag{2}
$$
and assume that $e$ is correlated with $z_1$. Then we can write:
$$
e = \gamma z_1 + \varepsilon.
$$
where $\varepsilon$ is now uncorrelated with $z_1$. (and where $\gamma = \mathbb{E}(e z_1)/\mathbb{E}((z_1)^2) \ne 0$. Assume for simplicity that $e$ is uncorrelated with $z_2$.
Then the estimate of $\beta_1$ will be biased as the orthogonality condition $\mathbb{E}(e x) = 0$ is not satisfied. Indeed:
$$
\begin{align*}
\mathbb{E}(ex) &= \mathbb{E}(e z_2) + \mathbb{E}(\gamma z_1 z_1) + \mathbb{E}(\varepsilon z_1),\\
&= \mathbb{E}(\gamma (z_1)^2) \ne 0
\end{align*}
$$
If we include $z_1$ into the regression. Then substituting $e = \gamma z_1 + \varepsilon$, into $(2)$ we can write:
$$
y = \beta_0 + \beta_1 x + \gamma z_1 + \varepsilon.
$$
And $\mathbb{E}(\varepsilon) = \mathbb{E}(\varepsilon x) = \mathbb{E}(\varepsilon z_1) = 0$. So by including $z_1$ into the regression, we can guarantee the residual $\varepsilon$ to be uncorrelated with all covariates. This means that $\beta_1$ is identified and its estimate will be unbiased.
