Linear projection results tell us that the OLS slope coefficients will, letting $x=(A\quad U')'$, tend to (there is no constant in the regression)
$$
\text{plim}\,\begin{pmatrix}\hat\alpha'\\\hat\beta'\end{pmatrix}=E(xx')^{-1}E(xy)
$$
We have
$$
E(xx')=\begin{pmatrix}E(A^2)&E(AU')\\E(AU')&E(U'^2)\end{pmatrix}
$$
and hence
$$
E(xx')^{-1}=\frac{1}{E(A^2)E(U'^2)-[E(AU')]^2}\begin{pmatrix}E(U'^2)&-E(AU')\\-E(AU')&E(A^2)\end{pmatrix}
$$
Also,
$$
E(xy)=\begin{pmatrix}E(A(\alpha A + \beta U)\\E(U'(\alpha A + \beta U))
\end{pmatrix}=\begin{pmatrix}\alpha E(A^2)+ \beta E(AU)\\\alpha E(U'A) + \beta E(U'U)
\end{pmatrix}
$$
so that
\begin{eqnarray*}
\text{plim}\,\hat\alpha'&=&\frac{E(U'^2)(\alpha E(A^2)+ \beta E(AU))-E(AU')(\alpha E(U'A) + \beta E(U'U))}{E(A^2)E(U'^2)-[E(AU')]^2}\\
&=&\alpha+\frac{E(U'^2)\beta E(AU)-E(AU')\beta E(U'U)}{E(A^2)E(U'^2)-[E(AU')]^2}
\end{eqnarray*}
If $\epsilon$ is independent of all other random variables (you only the state the marginal, but not the joint distribution), we have
$E(U'U)=E((U+\epsilon)U)=E(U^2)$ and similarly, $E(U'^2)=E(U^2)+1$ and $E(AU')=E(AU)$.
Thus,
\begin{eqnarray*}
\text{plim}\,\hat\alpha'&=&\alpha+\frac{\beta (E(U^2)+1)E(AU)-\beta E(AU)E(U^2)}{E(A^2)(E(U^2)+1)-[E(AU)]^2}
\end{eqnarray*}
I do not see any further "nice" simplification.
Illustration:
library(mvtnorm)
n <- 10000
cm <- matrix(c(1,0.5,0.5,1), ncol=2)
v <- rmvnorm(n, sigma = cm)
alpha <- 2
beta <- 3
epsilon <- rnorm(n)
A <- v[,1]
U <- v[,2]
y <- alphaA + betaU
U.prime <- U + epsilon
\alpha +\frac{\beta (E(U^2)+1)E(AU)-\beta E(AU)E(U^2)}{E(A^2)(E(U^2)+1)-[E(AU)]^2}
> alpha + (beta(cm[2,2]+1)cm[1,2]-betacm[1,2]cm[2,2])/(cm[1 .... [TRUNCATED]
[1] 2.857143
> lm(y~A+U.prime-1)
Call:
lm(formula = y ~ A + U.prime - 1)
Coefficients:
A U.prime
2.836 1.265