As pointed out in the comments, the difference between the models is that auto.arima() has not included an intercept. It selects a model, including the constant, using the AICc. With one covariate, the model is $$y_t = \beta_0 x_t + n_t$$ where $n_t$ is an ARIMA process. Note that the intercept is shifted to the ARIMA process. In this example, the selected model for $n_t$ does not include a constant.

If you know what model you want, why use auto.arima()? Instead, you could use

arima(a,xreg=b)

which gives

Series: a 
ARIMA(0,0,0) with non-zero mean 

Coefficients:
      intercept          b
       48638.40  -26143.23
s.e.   32410.27   27893.41

sigma^2 estimated as 93138232:  log likelihood=-254.25
AIC=514.5   AICc=515.7   BIC=518.03

This is the same as the model obtained using lm(a~b). The estimates are identical, but the standard errors are different because they are estimated in a different way (numerically from the hessian matrix rather than using the inverse of $(X'X)$.)

As pointed out in the comments, the difference between the models is that auto.arima() has not included an intercept. It selects a model, possibly including the constant, using the AICc. With one covariate, the model is $$y_t = \beta_0 x_t + n_t$$ where $n_t$ is an ARIMA process. Note that the intercept is shifted to the ARIMA process. In this example, the selected model for $n_t$ does not include a constant.

If you know what model you want, why use auto.arima()? Instead, you could use

arima(a,xreg=b)

which gives

Series: a 
ARIMA(0,0,0) with non-zero mean 

Coefficients:
      intercept          b
       48638.40  -26143.23
s.e.   32410.27   27893.41

sigma^2 estimated as 93138232:  log likelihood=-254.25
AIC=514.5   AICc=515.7   BIC=518.03

This is the same as the model obtained using lm(a~b). The estimates are identical, but the standard errors are different because they are estimated in a different way (numerically from the hessian matrix rather than using the inverse of $(X'X)$.)