Omitted Variables and their consequences for other variables Suppose we have the following regression model:
$$
y_{i}=\boldsymbol{x_{i}'\beta}+e_{i}
$$
where the vector $\boldsymbol{x_{i}'}$ contains two variables, $[x_{1i}\,x_{2i}].$Suppose
for a second, that $e_{i}$ contains variables that correlated with
$x_{1i}$. The well known omitted variables bias tells us that $\beta_{1}$
will be measured with bias when estimated by OLS. However, what implications
does this have for $\beta_{2}?$ In other words, does bias in one
set of estimates still have implications for bias in a second set
of estimates? In other words, if $cov\left(x_{1i},e_{i}\right)\neq0$
, but $cov\left(x_{2i},e_{i}\right)=0,$ what does this mean for the
estimates of $\beta_{2}?$
 A: Intuitively, the answer is that $\hat{\beta}_2$ will be biased whenever $x_2$ is correlated with $x_1$.  This is because the correlation of $x_2$ with $x_1$ causes $\hat{\beta}_2$ to be correlated with $\hat{\beta}_1$; as a consequence, when $\hat{\beta}_1$ differs from $\beta_1$, $\hat{\beta}_2$ will differ from $\beta_2$ in expectation, and, writing loosely, since we expect $\hat{\beta}_1$ to be biased, $\hat{\beta}_2$ will be too.
I'll assume away the intercept term for simplicity of exposition, without loss of generality.  Writing out the expression for $\mathbb{E}\hat{\beta}$ gives us:
$$\mathbb{E}\hat{\beta} = \mathbb{E}(X^TX)^{-1}X^TY = \mathbb{E}(X^TX)^{-1}X^T(X\beta + e) = \beta + \mathbb{E}(X^TX)^{-1}X^Te$$
Under the assumptions of the question, $\mathbb{E}X^Te$ can be written as $(\theta,0)$, where $\theta$ is a function of the nonzero covariance between $x_1$ and $e$.  Substituting gives us:
$$  \mathbb{E}\hat{\beta} - \beta = [(X^TX)^{-1}_{1,1}\theta, (X^TX)^{-1}_{2,1}\theta]$$
The subscripts designate which element of the $2 \times 2$ matrix $(X^TX)^{-1}$ is being referred to.
Since $(X^TX)^{-1}_{1,1} > 0$ unless $x_1$ is identically equal to $0$, we can see that $\hat{\beta}_1$ will be biased; however, $\hat{\beta}_2$ will be biased if and only if $(X^TX)^{-1}_{2,1} \ne 0$, i.e., if $x_1$ and $x_2$ are correlated.
Here's a simple example in R, where the true values of $\beta_1$ and $\beta_2$ are both equal to $1$:
bhat_1 <- rep(0, 10000)

V <- matrix(c(1,0.6,0.6,1), 2, 2)

for (i in 1:length(bhat_1)) {
  x <- rmvn(100, c(0,0), V)
  y <- x[,1] + x[,2] + rnorm(100)
  bhat_1[i] <- coef(lm(y~x[,1]))[2]
}


hist(bhat_1)

that produces the following histogram:

A: Here's what happens for any number of predictors and omitted variables. The answer ends up similar to what @jbowman (+1) showed for two predictors (one correlated with omitted variables and one not). Since I'd already started writing this, I'll post it in case it's useful to see another approach.
To summarize the answer below: Suppose $X_u$ contains predictors that are uncorrelated with all omitted variables and $X_c$ contains predictors that are correlated with at least one omitted variable. Then, under standard assumptions, OLS coefficients for predictors in $X_u$ are unbiased if and only if $X_u$ and $X_c$ are uncorrelated.
Setup
Let $y \in \mathbb{R}^n$ be a vector of responses and $X \in \mathbb{R}^{n \times p}$ be a matrix of predictors. We also have predictors $Z \in \mathbb{R}^{n \times q}$ that will be omitted. Assume the predictors and responses have been centered, so there's no need for an intercept term. I'll also assume that predictors may be correlated but not perfectly collinear (this implies that $X$ and $Z$ have full rank, and $n \ge p+q$).
Suppose true model is:
$$y = X w + Z v + \epsilon \quad \quad
\epsilon \sim \mathcal{N}(\vec{0}, \sigma^2 I)$$
where $w \in \mathbb{R}^p$ and $v \in \mathbb{R}^q$ are the true coefficients, and $\epsilon$ is a random vector representing i.i.d. Gaussian noise with mean zero and variance $\sigma^2$.
Say we fit an ordinary least squares regression model, omitting $Z$. The estimated coefficients are:
$$\hat{w} = (X^T X)^{-1} X^T y$$
The bias is a vector containing the expected difference between the estimated and true coefficients (see derivation below):
$$\text{bias}
= E[\hat{w} - w]
= (X^T X)^{-1} X^T Z v$$
Multiplying both sides by $X^T X$ gives:
$$X^T X \ \text{bias} = X^T Z v$$
For predictors uncorrelated with the omitted variables
Suppose the predictors are partitioned as $X = [X_u, X_c]$, where $X_u$ contains columns uncorrelated with all omitted variables and $X_c$ contains columns correlated with at least one omitted variable. So, $X_u^T Z = \mathbf{0}$ and $X_c^T Z \ne \mathbf{0}$. Similarly, suppose the bias is partitioned into subvectors $\text{bias}_u$ (for predictors in $X_u$) and $\text{bias}_c$ (for predictors in $X_c$). Rewrite the preceding equation in the partitioned form:
$$\begin{bmatrix}
  X_u^T X_u & X_u^T X_c \\
  X_c^T X_u & X_c^T X_c
\end{bmatrix}
\begin{bmatrix} \text{bias}_u \\ \text{bias}_c \end{bmatrix}
= \begin{bmatrix} \mathbf{0} \\ X_c^T Z \end{bmatrix} v$$
Break this into two systems:
$$X_u^T X_u \text{bias}_u + X_u^T X_c \text{bias}_c = \vec{0}$$
$$X_c^T X_u \text{bias}_u + X_c^T X_c \text{bias}_c = X_c^T Z v$$
$\text{bias}_c$ is nonzero, since assuming it's zero leads to a contradiction, given our assumptions. This recapitulates the standard statement about omitted variable bias.
More interestingly, the question concerns $\text{bias}_u$, the bias of predictors that are uncorrelated with the omitted variables. The first equation in the above pair leads to two conclusions: 1) If $X_u$ and $X_c$ are uncorrelated so $X_u^T X_c = \mathbf{0}$, then the only solution is $\text{bias}_u = \vec{0}$. Recall that $X$ has full rank so $X_u$ does too, and the null space of $X_u^T X_u$ includes only the zero vector. 2) Since $\text{bias}_c$ is nonzero, $\text{bias}_u = \vec{0}$ would imply that $X_u^T X_c = \mathbf{0}$.
Therefore, coefficients for predictors in $X_u$ have zero bias if and only if $X_u$ and $X_c$ are uncorrelated:
$$\text{bias}_u = \vec{0} \ \iff \ X_u^T X_c = \mathbf{0}$$
Derivation of the bias
The bias is the expected difference between the estimated and true coefficients, where the expectation is taken w.r.t. $\epsilon$:
$$\begin{array}{ccl}
  \text{bias} & = & E[\hat{w} - w] \\
  & = & E\big[ (X^T X)^{-1} X^T y - w \big] \\
  & = & E\big[
    (X^T X)^{-1} X^T X w
    + (X^T X)^{-1} X^T Z v
    + (X^T X)^{-1} X^T \epsilon
    - w
  \big] \\
  & = & E\big[ (X^T X)^{-1} X^T Z v + (X^T X)^{-1} X^T \epsilon \big] \\
  & = & (X^T X)^{-1} X^T Z v
  + E\big[ (X^T X)^{-1} X^T \epsilon \big] \\
  & = & (X^T X)^{-1} X^T Z v \\
\end{array}$$
Line 2 substitutes in the expression for $\hat{w}$. Line 3 substitutes the true model in for $y$. Line 4 is an algebraic simplification. Line 5 uses linearity of expectation. Since noise has zero mean and is uncorrelated with the predictors, the expectation in the last term is zero, giving line 6.
